Calculus
Real calculus questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
The equation has exactly three roots: x ≈ 1.28378, 2.49096 and 3.58430. Beyond x = 1 + e^1.5 ≈ 5.48 the logarithm term exceeds 3, so no further crossings can occur.
The solutions are x(t) = 1/(C - 2t^2) together with the constant solution x = 0, which separation loses when you divide by x^2. Full derivation and check.
The antiderivative is x/2 + sin(2x)/4 + C. Rewriting cos^2 x as (1 + cos 2x)/2 removes the square, leaving two elementary integrals instead of a product.
The antiderivative is D(Ka)^(−1/n) ₂F₁(1/n, 1/n; 1 + 1/n; −Dⁿ/Ka) + C. Pull out Ka, expand the binomial series and integrate term by term; n = 2 collapses to arcsinh.
The result is x^(1−b)[(1−b)sin(a ln x) − a cos(a ln x)]/((1−b)² + a²) + C. The substitution u = ln x turns it into a standard exponential-times-sine integral.
F(x) = (1/2) ln(2x + 1) and F((e^2 - 1)/2) = 1. Substitute u = 2x + 1, use F(0) = 0 to kill the constant, then note that 2x + 1 collapses to e^2.
The curves meet at x = -1 and x = 2, and integrating (x + 2) - x^2 across that interval gives an enclosed area of exactly 9/2 square units.
The area is 9/2. Find the intersections at x = 0 and x = 3, set up the double integral of 1 with y running from x up to 4x - x^2, then integrate the strip height 3x - x^2.
The region between x² + y² = 7y and x² + y² = 9y cut by y = x/√3 and x = 0 has area 8π/3 + 2√3 ≈ 11.84. Set up and evaluate the polar area integral step by step.
The average rate of change is 49. Compute (f(5) - f(3))/(5 - 3) = (125 - 27)/2, the slope of the secant line joining the two points on the cubic.
The sum is pi^2/6, about 1.644934. It converges as a p-series with p = 2, and Euler found the exact value by factoring sin(x)/x over its roots.
The solution is y^3 = 1/(1 + C(1+t^2)). The substitution v = y^-3 turns this Bernoulli equation into a first-order linear ODE with integrating factor 1/(1+t^2).
Answer: sum of (-1)^n C(k,n) x^n = 1 - kx + k(k-1)x^2/2! - ..., valid for |x| < 1. Each minus sign comes from substituting -x into the binomial expansion.
x = 1 is removable, a hole at (1, -2), while x = 2 is non-removable, a vertical asymptote. Factor top and bottom and see which zero factor cancels.
The sum equals e^(1 + 2/n)(e^2 - 1)/(e^(2/n) - 1). Each term is the previous one times e^(2/n), so the finite geometric formula applies directly.
It converges by comparison with the Bertrand series 1/(n ln^2 n). There is no elementary closed form; the sum is about 2.4919, not the 2.42 a short partial sum suggests.
Absolutely convergent for a > 1, conditionally convergent for 1/2 < a <= 1, divergent for 0 < a <= 1/2 - the second-order term of ln(1+x) decides the borderline case.
The integral converges absolutely for p > 2, converges conditionally for 1 < p <= 2 and diverges for p <= 1. Solved by the substitution u = x + y plus Dirichlet test.
The curve never rises above the x-axis, has a minimum at (-2, -8/e^2) about -1.083, a maximum at the origin, and tends to 0 from below as x goes to minus infinity.
The answer is 2 pi, about 6.2832. Integrating the constant 1 measures the length of the interval, so the result is simply the upper limit minus the lower.
The answer is 876.93758, not 876.93843. The decay term contributes a*k = 20.67158 because e^(-23.58) is negligible, and the constant contributes 856.266.
The value is (7 − π − 3√3 − 2√2)/384 ≈ −0.010849. It is negative only because the lower limit π/8 exceeds the upper limit π/12, which flips the sign.
Answer: pi/6, about 0.5235988. The substitution u = x^3 turns the integral into a Beta function B(1/2, 3/2)/3, which Gamma values collapse to pi/6 exactly.
The definite integral equals 8. Antidifferentiate x to x^2/2, evaluate at 4 and 0, and confirm with the area of the triangle under the line y = x.
Answer: 10/3, about 3.3333. Integrate term by term to get x^3/3 + 3x, then apply the fundamental theorem: the value at 1 is 10/3 and the value at 0 is 0.
The value is 4 sin 2 - 2 cos 2 - 2, about 2.4695. With limits that are not multiples of pi, the sine and cosine terms survive instead of collapsing.
The value is pi^2 - 4, about 5.8696. Integration by parts with u = x^2 drops the power to x, and a second pass finishes it; the sine terms vanish at both limits.
The integral equals exactly pi. The x^3 cos(x/2) piece is odd, so it integrates to 0, and the rest is half the area of a circle of radius 2.
The exponential solutions are f(x) = C e^(rx) where r e^r = 1, so r = W(1) is about 0.567143. The unit delay makes the characteristic equation transcendental.
The derivative is 3x² − 6x + 2 and the integral from 0 to 3 equals 9/4. Power rule first, then the Fundamental Theorem of Calculus for the signed area.
The domain is all real numbers and the derivative is y' = -2sin(x). The constant 2 passes straight through differentiation, and cosine differentiates to minus sine.
y' = 30x^2 - 2x + 4, and y' = 0 has no real solution because its discriminant is 1 - 120 = -119 < 0, so the curve has no horizontal tangent anywhere.
The derivative is 6e^(3x). The exponential reproduces itself, the chain rule contributes the inner derivative 3, and the constant 2 rides along untouched.
The answer is 6x^2. Expanding (x+h)^3, cancelling the constant term and dividing by h leaves 6x^2 + 6xh + 2h^2, whose limit as h approaches 0 is 6x^2.
The derivative is h′(x) = 2(2x + 1)⁴(x² − 1)²(11x² + 3x − 5). Apply the product rule, use the chain rule on each power, then factor out the common (2x + 1)⁴(x² − 1)².
dy/dx = 36x^2 cos(4x^3+5) sqrt(5(4x+5/x)) + 3 sin(4x^3+5)(20 - 25/x^2)/(2 sqrt(5(4x+5/x))). Found with the product rule applied to two chain-rule factors.
The answer is dy/dx = -30x sin(3x^2). The outer cosine contributes -sin of the same inside, and the inner 3x^2 contributes a factor of 6x on top of the 5.
Answer: dy/da = -e^(-a/2)/(2(1 + e^(-a))) = -e^(a/2)/(2(e^a + 1)). Note that squaring e^(-a/2) gives e^(-a), which is what makes the denominator collapse.
The derivative is -sin(sqrt(2x+1))/sqrt(2x+1) for x > -1/2. The chain rule applies twice, and the inner factor 2 cancels the 1/2 from the power rule.
The answer is (e^x(x - 3) - 3)/x^4. Rewriting the fraction as (e^x + 1)x^(-3) turns a quotient rule problem into a shorter product rule one.
The derivative is 1/(x + e*pi). The chain rule gives (1/e*pi) over (1 + x/(e*pi)), and multiplying top and bottom by e*pi clears the nested fraction.
The derivative is 2/x. The chain rule gives 6x/(3x^2) and the constant 3 cancels, exactly as log rules predict, since ln(3x^2) = ln 3 + 2 ln x.
The derivative of y = log(sin x) is cot x. Apply the chain rule with the inner function sin x, then simplify cos x over sin x into a single cotangent.
Answer: y prime = x*cos(3.14/3 - x^2). The inner derivative is -2x, and the two minus signs cancel against the -1/2, leaving a clean positive coefficient of x.
y' = 3x^11 + 18x^5 - 1. The fractional coefficient 1/4 multiplies the exponent 12 to give 3, and the constant term 10 contributes nothing to the derivative.
y' = (1/2)x^4 + 4x^3 + 3. The leading coefficient becomes 5/10 = 1/2, not 2, and the irrational constant sqrt(3) differentiates to zero like any constant.
The derivative is 2 sin x cos x, which the double-angle identity rewrites as sin 2x. Read sin^2 x as (sin x)^2 so the outer function is the square.
The derivative is 3x² − 6x + 2. The power rule brings each exponent down as a coefficient and drops it by one, so x³ → 3x², −3x² → −6x, and 2x → 2.
The derivative is 3x² + 4x − 5. Apply the power rule term by term, keep the constant multiples, and note that the derivative of a linear term is just its coefficient.
The derivative is (x - 1)^x [ln(x - 1) + x/(x - 1)] for x > 1. Neither the power rule nor the exponential rule applies, so take ln of both sides first.
The derivative is (1 − x²)/(x² + 1)². Take u = x and v = x² + 1, so u′ = 1 and v′ = 2x, then (u′v − uv′)/v² collapses to (x² + 1 − 2x²)/(x² + 1)².
The answer is (1 - x^2)/(x^2 + 1)^2. The numerator collapses from (x^2+1) - 2x^2, which is why the derivative changes sign exactly at x = 1 and x = -1.
The derivative is (2x tan x - x^2 sec^2 x)/tan^2 x. Apply the quotient rule with d/dx(tan x) = sec^2 x, then rewrite the result with sine and cosine.
The derivative of f(x) = x^2/2 + cos 5x is x - 5 sin 5x. The quadratic term gives x, and the chain rule on cos 5x pulls out a factor of 5 with a sign flip.
The answer is 2x. The power rule d/dx x^n = n*x^(n-1) with n = 2 gives 2x^1, and the limit definition confirms it - the 2xh term is what survives.
Eliminating a and b from x²/a² − y²/b² = 1 gives xy y″ + x(y′)² − y y′ = 0. Differentiate twice, then divide the two relations to remove both constants at once.
dy/dx = 40x^4 + 12/x^5 - 4/x^2. Each term is written as a power of x first, so the power rule applies uniformly and the two negative exponents flip signs.
dy/dx = 48x^5 + 12/x^5 - 4/x^2. Rewriting 4/x as 4x^-1 lets the power rule handle every term, and the negative exponents flip two signs along the way.
The derivative is 2e^(2x) artanh(sqrt(e^x-1)) + e^(3x)/(2 sqrt(e^x-1)(2-e^x)). Note 1 - z^2 collapses to 2 - e^x, which also fixes the domain 0 < x < ln 2.
The answer is (3 - e^2x)/(e^x + e^-x)^3. Treating the denominator as a negative power turns the quotient rule into a cleaner product-and-chain-rule calculation.
dy/dx = 2x. The power rule d/dx of x^n equals n times x^(n-1), so with n = 2 the old exponent becomes the coefficient and the new exponent drops to 1.
Integrating v(t) = t^2 + t from t = 1 to t = 2 gives a displacement of 23/6 metres, about 3.83 m, and the velocity never changes sign on that interval.
Answer: about -33.0784. The integrand is free of x, so the inner integral is just its width y/1.1015, which is negative for y in [-1.65, 0] - hence the negative value.
The value is 72. The three lines meet at (-2,-2), (4,-2) and (4,4), so integrating y from -2 up to x and then x from -2 to 4 covers the triangle exactly.
The answer is −8. The inner limits run from y = x to y = 3, so the inner integral leaves 3x² − 8x − 3, which integrates to −8 over 1 ≤ x ≤ 2.
The answer is 3/32 + ln(2)/8 ≈ 0.180393. The hyperbola xy = 1/4 splits the square, and the 1/(16x) piece is what produces the logarithm in the final value.
The answer is π(1 − 1/e) ≈ 1.9858. In polar coordinates the integrand becomes e^(−r²)·r, whose r-antiderivative exists in closed form even though e^(−x²) has none.
The exact value is (pi/4)erf(1)² ≈ 0.55775. The integrand factors, so the square integral becomes the square of a single Gaussian integral — no elementary antiderivative needed.
The value is the sum of two Fresnel integrals, about 1.2148. Neither piece has an elementary antiderivative, but the sum splits into two one-variable integrals.
The integral equals 5/6. The substitution u = 1 - x, v = 1 - y turns the max into max(sqrt(1 - u^2), v), splitting the square along a quarter circle.
Answer: -3pi/2. Integrating in y from 0 to x gives x(sin 2x - sin x), and the two resulting integrals over [0,pi] are -pi/2 and pi.
The value is -16. On the symmetric interval -1 <= x <= 1 the x term integrates to zero, so only -4y survives and the answer comes from a single one-variable integral.
The value is -9/2. Integrate x - 4y in x from -1 to 2, then in y from 0 to 1; reversing the order of integration gives the same -9/2, as Fubini's theorem guarantees.
The value is 5/3. Distribute the x/8 first so the integrand becomes a sum of monomials, then integrate in y and x — the constant 1/8 makes this a joint-density integral.
The integral equals 1 - root 2 over 2, about 0.292893. Reversing the order splits the region into two y-slices; polar coordinates confirm the same value in one line.
The integral equals ln 2. Integrating in y gives an arctangent, the bounds collapse to arctan 1 minus arctan(x/2), and one integration by parts finishes it.
The value is (7/6)(√2 − 2) ≈ −0.6834. Convert to polar, intersect the conditions x ≤ 0 and y ≥ x to get the wedge from pi/2 to 5pi/4, then integrate r² and the angles.
Answer: pi/6 - root3/8 + 5/48, about 0.411259. Factor x(1 + y), integrate in x to get root(1-y^2) - 1/2, then use the standard arcsine antiderivative.
The value is 4pi/sqrt(3) - 3pi/2, about 2.5428. In polar form the r cancels, leaving a single theta integral that the sum-of-cubes identity collapses.
Answer: -5/3. The integrand splits into x^2 and -8y, each of which separates over the rectangle, giving 7/3 - 4 = -5/3.
Answer: 2/3. By symmetry the two terms contribute equally, each giving 1/3, so the integral over 0<=x<=1, 0<=y<=1 is 2/3.
The value is pi/2 − sin(pi³)/(2pi²) ≈ 1.59097. The inner integral is a perfect u-substitution: u = xy² makes xy dy exactly du/2, collapsing sin(xy²) to a cosine.
Answer: 5pi/4 + 1/8 - e^(10pi)/8. The substitution u = x^2 y turns the inner x-integral into (1 - e^(4y))/2, and the outer integral is elementary.
The answer is 4/5. Vertical slices run from y = x² up to y = 1 for −1 ≤ x ≤ 1, giving (1 − x⁴)/2 after the inner integral and 4/5 after the outer one.
Answer: 0. The integrand separates, and the x-factor x/root(R^2-x^2) is odd on the symmetric interval [-R, R]; each half converges to R, so the sum is exactly zero.
Compute the double integral of x squared minus 8y over the rectangle -2 <= x <= -1, 0 <= y <= 1. Step-by-step iterated integration with constant limits.
The double integral equals 16pi/9 = 5.585. Integrate in y first to get (8/3)x sin^3 x, rewrite sin^3 x with the triple-angle identity, then use parts.
Evaluate the double integral of x cos(x + y) over the triangle (0,0), (π,0), (π,π). Limits of 0 ≤ y ≤ x give the exact answer −3π/2 ≈ −4.712.
Evaluate the double integral of 9y + 4x over the region bounded by y = 4x² − 9 and y = 9 − 4x². Find the limits, then use symmetry to get the answer instantly.
Answer: sin^2(x)/128 + (1 - cos x)/72 for 0 <= x <= pi/2. Factor r^2 sin(theta)(r cos theta + 1/3) to show the integrand is nonnegative, then drop the absolute value.
The value is about 1.7865. A plus sign inside the root means the modulus is imaginary, so E(k) is used with a negative parameter and evaluated numerically.
The value is about 1.928009. The integral is the complete elliptic integral of the second kind E(m) with the negative parameter m = -tan^2(0.8), which is about -1.06016.
Both ends tend to 0 because the denominator has degree 6 and the numerator only degree 3. The quotient behaves like -1/(2x^3), giving a horizontal asymptote y = 0.
The series 1/Gamma(3/5) times the sum of (1/10)^(8n/5+1)/Gamma(8n/5+1) equals about 0.00118532. Only the first three terms matter; the rest fall below 1e-10.
The general solution is x + 2y − 3ln|x + y + 4| = C. The equation is not exact, but both coefficients depend only on x + y, so u = x + y separates the variables.
Answer: a = 2, b = 3, c = 4. The numerator must vanish at x = 1 for a finite limit, factoring gives limit 3, and the left branch −x + c must match it.
The limit exists only when a = e - 1/e. The factor with e^(1/x) tends to 1 from the right and 2 from the left, while (1+|x|)^(1/x) tends to e and 1/e.
The general solution is y = e^(−x) + Ce^(−2x). Multiply through by the integrating factor e^(2x), recognise the product rule, integrate once and divide back.
The surface integral equals 8 pi. Cap the open hemisphere with the disk z = 0, apply the divergence theorem, subtract the disk flux, then flip the sign for the inner side.
The sine term dies by odd symmetry, the cosine term is the 3-D Gaussian Fourier transform, and the r² term is a gamma integral, leaving two clean closed-form pieces.
The answer splits by sign: k > 0 gives cos and sin of root k times t, k = 0 gives a straight line C1t + C2, and k < 0 gives growing and decaying exponentials.
For f = x^2 y + x e^(-xy) the directional derivative in the direction (0, -1) is -x^2(1 - e^(-xy)) - it is just minus the partial derivative with respect to y.
dy/dx = -(3x^2 + 2xy)/(x^2 + 3y^2). Differentiate every term in x, use the product rule on x^2*y and the chain rule on y^3, then solve for dy/dx.
dy/dx = (-180x^5 - 400x^19 y)/(20x^20 + 8y^7). The term 20x^20 y needs the product rule, and every y differentiated contributes a dy/dx factor.
The value is 1/3. The substitution x = cosh t turns x + sqrt(x^2-1) into e^t, and the lower limit must be 1, not 0, for the integrand to be real.
The answer is −1. The antiderivative x ln x − x comes from integration by parts, and the endpoint at 0 is handled by the limit a ln a → 0 as a → 0⁺.
The integral from minus infinity to 0 of 1/(3 root of (x-8)^2) diverges to plus infinity. Rewrite the root as 8 - x, integrate to a logarithm, and take the limit.
The value is exactly 1/2. Substituting u = x² turns the integrand into (1/2)e^(−u), and letting the upper bound grow makes e^(−b²) vanish, so the integral converges.
The value is 6. The integrand blows up at x = 0, so the integral must be split there, but each half converges to 3 because the antiderivative 3x^(1/3) stays finite.
The answer is -(1/b)ln|a - bx| + C. The substitution u = a - bx has du = -b dx, so the reciprocal of the inner slope, negative included, comes out front.
The answer is 2ln|x| + 1/x + C. The exponent -1 needs the logarithm while -2 uses the ordinary power rule, so the two terms behave quite differently.
The answer is 3ln|x| + 2/x + C. The x^-1 term needs the logarithm exception while x^-2 uses the ordinary power rule, and the double negative turns into a plus.
The answer is Ei(x) + C, the exponential integral. No combination of elementary functions differentiates to e^x/x, so a special function is genuinely required.
The antiderivative is -(1/2)e^(-x^2) + C. The factor x outside is exactly half of du, which is what makes this substitution close perfectly.
The product converges for every x > 0 by comparison with the sum of 1/n^2, and ln of the product divided by root x tends to pi, which the Euler sinh formula makes exact.
Only two: m = −1 and m = 0. The derivative (4 − m²)/(x − m)² needs m² < 4, and the vertical asymptote x = m must stay outside (0, ∞), forcing −2 < m ≤ 0.
The answer is (b − a)u/(ab√(a − bu²)) + arcsin(u√(b/a))/(b√b) + C. Rewriting u² as (a − (a − bu²))/b splits the integral into two standard forms.
The answer is 2 arsinh(root x) + C, equivalently 2 ln(root x + root(x+1)) + C. The substitution x = sinh^2(t) makes both radicals rational at once.
The antiderivative is ln|x| + C. Without the absolute value the formula would be undefined for x < 0, where 1/x is perfectly well behaved and integrable.
The antiderivative is −1/x − (2/3)√x + C. Rewrite both terms as powers, x^(−2) and (1/3)x^(−1/2), then add one to each exponent and divide by it.
The antiderivative is ln|(1 + x)/(1 - x)| + C. Since 1 - x^2 = (1 - x)(1 + x), the integrand splits into 1/(1 - x) + 1/(1 + x) and the two logs combine.
The integral of 25x^(3/2) is 10x^(5/2) + C. Add 1 to the exponent 3/2 to get 5/2, then note that dividing by 5/2 is the same as multiplying by 2/5.
The answer is 2e^(2x) + (1/2)cos(2x) + C. Each inner coefficient 2 divides the result, and the minus in front of sin 2x makes the cosine term positive.
The antiderivative is 2e^(2x) + (1/2)cos 2x + C. Each term divides by the inner derivative 2, and the minus sign in front of sin 2x makes the cosine term positive.
The value is about 1.09869. There is no elementary closed form, so 1/cosh x is expanded as an alternating exponential series and the result confirmed by quadrature.
The answer is (1/2)sin(2x) + C. The chain rule would multiply by 2 on the way back, so the antiderivative must carry the reciprocal 1/2 in front.
The integral diverges as an ordinary improper integral; as a distribution it equals 2πδ(x). Truncating at ±A gives 2sin(Ax)/x, which oscillates forever but converges weakly to 2πδ(x).
The result is (sqrt(pi)/2)erf(x) + C. No elementary antiderivative exists, so the integral is defined through erf, whose 2/sqrt(pi) factor cancels here.
The value is 12*zeta(3)^2 - 2*pi^6/45, about -25.3891. Split the kernel, expand in series, and evaluate the Euler sum of H(m)/m^5 in closed form.
Integration by parts leaves the non-elementary piece ln(1+x^2)/x, which equals -Li2(-x^2)/2. The complete antiderivative therefore needs the dilogarithm function.
There is no elementary antiderivative. Rescaling to the standard form gives -2*sqrt(10*pi)*C(sqrt(8/(5*pi))*t) + C1, where C(x) is the Fresnel cosine integral.
No elementary antiderivative exists. Substituting u = ln x and expanding 1/(1 + e^(−u)) gives (ln x)^(1−a)/(1 − a) plus a series of gamma functions Γ(1 − a, n ln x).
Peel off one cosine, convert the rest with cos^2 = 1 - sin^2, then substitute u = sin x to obtain sin^3(x)/3 - sin^5(x)/5 + C as the antiderivative.
The answer is (1/4)cot(4 cos x) + C. The substitution u = 4 cos x turns the integrand into -(1/4)csc^2(u), whose antiderivative is (1/4)cot(u).
The answer is -cos(7x^4 + 3)/28 + C. The x^3 outside is almost the derivative of the inside 7x^4 + 3, so substitution works and integration by parts is unnecessary.
The antiderivative is -1/(2(x^2-2x+3)) + C. The numerator is half the derivative of the base, which is exactly the pattern u-substitution is built for.
The antiderivative is −5 ln|x − 2| + 6 ln|x − 3| + C. Factor the denominator into (x − 2)(x − 3), split into two simple fractions, then integrate each logarithm.
Answer: -(x/2)e^(-x^2) + (sqrt(pi)/4)erf(x) + C. Pair one x with the exponential so the parts step works, then express the leftover Gaussian with erf.
The answer is e^x (x^2 - 2x + 2) + C. See how choosing u = x^2 lowers the power each round, why two passes are needed, and how to check by differentiating.
Answer: -x^2 cos x + 2x sin x + 2 cos x + C. Each round of parts lowers the power of x by one, so a degree-2 polynomial needs exactly two applications.
The answer is -3pi/2. Split into two by-parts integrals: x sin(2x) gives -pi/2 and x sin(x) gives pi, so subtracting leaves -pi/2 - pi, which is -3pi/2.
The answer is (1/5)(x^2+1)^(5/2) - (2/3)(x^2+1)^(3/2) + (x^2+1)^(1/2) + C. Writing x^4 as (u-1)^2 turns the integrand into three simple powers of u.
The antiderivative is x^4/4 - 3x^2/2 + 5 ln|x| + C. Divide every numerator term by x first, then apply the power rule and the 1/x log rule separately.
The antiderivative is (25/7) y^(7/5) + C. Adding 1 to the exponent 2/5 gives 7/5, and dividing by 7/5 means multiplying by 5/7, so 5 times 5/7 is 25/7.
The antiderivative is e^(6x)/6 + C. The chain rule puts a factor of 6 in front when you differentiate, so integrating must divide by 6 to compensate.
The antiderivative is x^4/4 + 6x + C. The power rule raises the exponent from 3 to 4 and divides by 4, while the constant 6 integrates to the linear term 6x.
The antiderivative is x^2/2 + 3x + C. Split the integral term by term, apply the power rule to x and treat 3 as the constant 3x^0, then add one constant C.
The interval is the closed [-1, 1]. The coefficient collapses to 2/(n(n+2)), which decays like 1/n^2, so the radius is 1 and both endpoints still converge.
The result is 0.4δ(t−1) + 0.2δ(t−2) + 0.4δ(t−3). A bare e^(−as) with no F(s) attached inverts to a shifted Dirac delta, and the sign of the exponent fixes the shift.
Answer: no. Li2 is defined and continuous at 1 with value pi^2/6, but its derivative -ln(1-x)/x diverges to +infinity as x approaches 1 from the left.
The answer is π/4 + 1/6. The x² terms cancel after expanding the square, leaving a quarter-circle area integral plus two elementary polynomial pieces.
The answer uses the secant-cubed formula. After the inner r integration the problem becomes the integral of sec^3, which needs the standard log-plus-product antiderivative.
Solution: y(t) = 2e^t − t² − 2 + 4u(t−2)sinh(t−2). Transform the impulse to e^(−2s), split the algebra with partial fractions, and invert term by term.
Solve y′(t) − 2∫₀ᵗ y(τ)sin(t−τ)dτ = 1 with y(0) = −1. The convolution theorem turns the integral into a product, giving y(t) = −1/2 − ½e^(−t/2)cos(√7t/2) + …
Answer: 1/(s-2)^2 for s > 2. Start from L{t} = 1/s^2 and apply the first shifting theorem, which replaces s by s - 2 whenever a factor e^(2t) is present.
The left sum is 55 and the right sum is 74.5 for f(x) = 2x^2 - x - 1 on [2, 5]. Because f is increasing there, they bracket the exact area of 64.5.
The limit is -infinity. Differentiating the series gives (e^x - 1)/x, so the sum is an integral that behaves like -ln|x| and drifts down without bound.
The limit of (1 - cos 2x)/(2x) as x approaches infinity is 0. Because 0 <= 1 - cos 2x <= 2, the squeeze theorem traps the quotient between 0 and 1/x.
The limit is 0. The numerator vanishes like x²/2 while the denominator vanishes only like −x, so the quotient behaves like −x/2 and tends to zero from both sides.
The limit is 2. Here the radical sits in the denominator, so multiply by its conjugate, use 1 - x^2 = (1-x)(1+x) and cancel the common factor.
The limit is 0. Squaring erases the sign, so x^2 tends to +infinity from either side, and a fixed numerator over an unbounded denominator collapses to zero.
Answer: 256. Cancel the common factor x - 2 so the base becomes (x^2+2x+8)/4, which tends to 4 rather than 1; with exponent 4 the limit is 4^4 = 256.
The limit is −1. Dividing top and bottom by the faster-growing 3ˣ leaves ((2/3)ˣ − 1)/((2/3)ˣ + 1), and (2/3)ˣ → 0 because the base is below 1.
The limit is 0. Dividing top and bottom by x^2, the highest power in the denominator, leaves (2/x - 3/x^2)/(1 + 3/x), whose numerator vanishes.
Answer: the limit does not exist. Factoring to (x-1)(2x+3)/|x-1| gives -5 from the left and +5 from the right, a jump discontinuity of height 10 at x = 1.
The limit is 7/13. Both numerator and denominator vanish at x = 1, so each contains the factor x - 1; cancelling it removes the indeterminate form.
The limit is 42. Every polynomial is continuous everywhere, so you may substitute x = 2 straight into 3x³ + 5x² + 4x − 10 and evaluate: 24 + 20 + 8 − 10 = 42.
The limit is 1. Factor the numerator as (2-x)(2+x) and the denominator as x^2(2+x), cancel the shared (2+x) and substitute x = -2 into (2-x)/x^2.
Answer: 14/15. Check the denominator at x = 2 first - it equals 30, not zero - so direct substitution is legal and gives 28/30, which reduces to 14/15.
The limit is -infinity. Subtracting the unbounded x^2 from the constant 9 drives the value down without bound, matching a downward-opening parabola.
Since 1/i = -i, the expression equals sin x - i cos x, which keeps modulus 1 and circles forever. It never settles, so the limit does not exist.
f(x) tends to 1/4 and f(f(x)) tends to 7/4. Split the fraction to get the inner limit, then use continuity to evaluate the outer function at that limit.
The limit equals 2 ln 2, about 1.386. One factor grows like x ln 2 and the other shrinks like 2/x, so the infinity times zero product balances exactly.
Answer: 1. The denominator |x-1| equals 3 at x = -2, not 0, so the function is continuous there and direct substitution gives 3/3 = 1 with no one-sided work.
For a(n) = the integral from 1/2 to 2 of 2^x/(1+x^2)^n, the ratio a(n+1)/a(n) is a weighted average of 1/(1+x^2), so it tends to its maximum 4/5.
The limit is 5. Only the leading degrees matter: the logarithms behave like 100 ln(x) and 20 ln(x), and the constants 100 and 2 wash out entirely.
Answer: minus infinity. The rational part tends to the finite value 3 while the square root grows like n, so a bounded quantity minus an unbounded one diverges.
The limit is 1/18. Both parts vanish at x = -5, so multiply by the conjugate of the numerator and factor the quadratic to cancel the common (x + 5) factor.
The limit is 1. Factoring the numerator in cos x and rewriting sin^2 x as (1-cos x)(1+cos x) cancels the vanishing factor exactly, with no L'Hopital needed.
Answer: 1. Factoring gives cos x (1 - cos^2 x) = cos x sin^2 x, so the quotient is cos x times (sin x / x)^2, and both standard limits equal 1 at x = 0.
Answer: 1. The form is 0/0, and the Taylor series, the definition of the derivative of e^x at 0, and L Hopital rule all give the same limit of exactly 1.
The limit is 1. It is the definition of the derivative of eˣ at 0, it follows from the series eˣ = 1 + x + x²/2 + …, and l’Hôpital gives it in one step.
The limit is −2. Subtracting 2x kills the linear term of ln(1 + 2x), so the quadratic term −(2x)²/2 is what survives and dividing by x² leaves exactly −2.
The limit is 0. A logarithm grows slower than any positive power, so L'Hopital turns ln(n)/n^2 into 1/(2n^2), which collapses to zero as n grows.
The limit is 0. Factoring e^x inside the log gives ln(e^x + x) = x + ln(1 + x e^-x), so the numerator grows like x while the denominator grows like x^2.
The limit is +infinity. A negative coefficient times a variable heading to -infinity gives a positive product, and the fixed -4 cannot tame unbounded growth.
The limit is 0. The numerator only alternates between -1 and 1 so it stays bounded, while n grows without bound, forcing the absolute value 1/n to zero.
Substituting u = 2/n turns the sequence into 4 times (1 - cos u)/u^2, and since that ratio tends to 1/2 the limit of the whole expression is 2.
The right-hand limit is 1. Taking logs turns infinity^0 into x^2 ln(1/x), which tends to 0; on the left the base is negative so the expression is undefined.
The limit equals 98/3 = 32.667. Learn how to read a, b and f(x) off a Riemann sum: 2/n gives the width, 3 + 2i/n gives the right endpoints on [3, 5].
The limit equals sqrt(3)/6, about 0.2887. Multiply by the conjugate to clear the radicals, cancel the x, then substitute x = 0 into what is left.
Answer: 3 root2/2, about 2.12132. Substitute t = root x to expose a derivative-quotient, or rationalise both radicals; either route gives the same value.
Because sqrt(x^2) equals the absolute value of x, the right-hand limit is 2 and the left-hand limit is -2, so the two-sided limit fails to exist here.
The limit does not exist. At x = 5 the numerator is √16 − 6 = −2, not 0, so this is −2/0 rather than 0/0: the quotient runs to −∞ from the right and +∞ from the left.
The limit is 1/6. Multiplying by the conjugate turns the numerator into x - 4, which cancels the denominator and leaves 1/(sqrt(x+5) + 3) to evaluate.
The limit is 5/8. Substituting x = pi/2 + t gives (cos 2t - cos 3t)/(4t^2), and the second-order cosine expansions leave (9/2 - 2)/4.
The limit is 2/3. Dividing numerator and denominator by x turns every term into a standard sin(kx)/x limit, giving (3 + 1)/(1 + 5) = 4/6.
The limit is -3/2. Shifting x = pi/4 + t turns the numerator into -sin 4t - sin 2t and the denominator into 4t, so the ratio tends to (-4 - 2)/4.
The limit of sin(9x)/cot(12x) as x approaches 0 equals 0. Rewrite cot(12x) as cos/sin so the quotient becomes sin(9x)sin(12x)/cos(12x), then substitute.
The answer is 0. Dividing by cot(12x) is multiplying by tan(12x), which turns the indeterminate-looking quotient into a product of two factors that both go to 0.
Answer: the limit does not exist. Two sequences running to infinity give sin x = 1 and sin x = -1 forever, so the values never settle on a single number.
Answer: 1, provided x is in radians. The squeeze theorem with cos x <= sin(x)/x <= 1 proves it, and in degrees the same limit would instead be pi/180.
The limit is e^(1/3) ≈ 1.3956. It is a 1^infinity form, so take logarithms and compare the x² terms of ln(sin x / x) and cos x − 1 to get the exponent 1/3.
Answer: root2/2, about 0.70711. Rewrite both hyperbolic functions with exponentials, factor out e^x from each, and the surviving constant is 1/root2.
The limit is 7/12. Multiplying by the conjugate turns the ∞ − ∞ form into (7x + 49)/(√(36x² + 7x + 49) + 6x), whose leading behaviour is 7x/12x.
Write tan 2x as sin 2x / cos 2x, divide numerator and denominator by x, and the standard sin(kx)/x limits give 2/3 since cos 0 equals 1.
The limit is 0 because tan x - sin x behaves like x^3/2, one order above x^2. Replacing both functions by x first would wrongly collapse the numerator to zero.
The limit is 1. It is a 1^infinity indeterminate form; taking logs turns it into (x^2/3)/x, which tends to 0, so the original tends to e^0.
The answer is 3. A constant function has the same output for every input, so no matter how large x grows the value never moves away from 3 at all.
The answer is 2. Multiplying the sum by 1/2 and subtracting telescopes the coefficients into a plain geometric series, giving S/2 = 1 and therefore S = 2.
The limit is 0. Rewriting as x^3/e^(2x) gives infinity over infinity, and each differentiation drops the power by one while the exponential survives unchanged.
The limit equals 6. Writing x - 9 as (sqrt(x) - 3)(sqrt(x) + 3) cancels the denominator directly, leaving sqrt(x) + 3, which is 6 when x = 9.
The limit is −8. Substitution gives 0/0, so factor x² − 16 as (x − 4)(x + 4), cancel the x + 4 against the denominator, and evaluate x − 4 at x = −4.
The limit is -root 3, about -1.732. Both parts vanish at x = 2, so factor the numerator and multiply by the conjugate to cancel the shared (x - 2) factor.
The limit is +infinity. The x^2 term outgrows every bound while the -4 stays fixed, so the difference diverges upward and no finite limiting value exists.
The limit is 4/3. Direct substitution gives 0/0, so factor both parts into (x + 5)(x − 3) and (x − 3)(x + 3), cancel the shared x − 3, then substitute x = 3.
The limit of (x^2 + 5x + 4)/(x^2 + 3x - 4) as x approaches -4 is 3/5. Both parts vanish at -4, so factor out the shared (x + 4) before substituting.
Answer: the limit does not exist. Factoring gives (x-3)(x+2)/|x+2|, whose left limit is 5 and right limit is -5, so the two one-sided values disagree.
The limit is −25. Only the linear terms matter: the numerator behaves like 5x and the denominator like −x/5, because the √(1−x²) part contributes nothing at first order.
The limit equals 5/16. Two ways to get there: split the numerator into two conjugate pieces, or apply L'Hopital's rule once to the 0/0 indeterminate form.
The answer is 16/3 + 2 sin 2. Split off the exact differential and evaluate it from the endpoints, then parametrise the rest with x = 2cos t, y = 2sin t.
Answer: ln(x/y) is about (x-y)/y, the relative change. Adding the next series term gives (x-y)/y - ((x-y)/y)^2/2, which is accurate to well under a percent.
Linearising f(x) = (64 + x)^(1/3) at x = 0 gives L(x) = 4 + x/48, so the cube root of 62.5 is about 3.96875 against the exact 3.968503 — an error of 0.00025.
The value is (π/sin(π/10))(ψ(0.1) + γ) ≈ −100.103. Differentiating B(a, b) in a must keep the Γ(a+b) in the denominator, or the answer comes out about 6% too negative.
The series is 1 − 2x + 3x² − 4x³ + 5x⁴ − 6x⁵ + 7x⁶ − 8x⁷ + …, that is Σ(−1)ⁿ(n + 1)xⁿ. It comes from differentiating the geometric series for 1/(1 + x).
The series is x − x³/6 + x⁵/120 − x⁷/5040. Only odd powers appear because sine is odd, and the denominators are the odd factorials 3!, 5!, 7!.
The series is x − (2/3)x³ + (4/5)x⁵ − (47/63)x⁷ + O(x⁹). Multiply the tangent series by the geometric series for 1/(1 + x²) and collect powers up to x⁷.
The answer is -k. Expand A x B = 3yz i - 4xz j - xy k, differentiate once in x to get -4z j - y k, then once in y, which leaves the constant vector -k.
After 20 minutes the coffee is 48.9 C, and it reaches 30 C after about 44 minutes. Fit k = (1/10)ln(14/9) = 0.04418 from the 65 C reading at 10 minutes.
From the right the limit is 0 and from the left it is +infinity, so the two-sided limit does not exist. The substitution t = 1/x turns each side into 4t^2/2^t.
The answer is 36 pi^2 < a < 64 pi^2. Each k with 4 pi^2 k^2 < a contributes a pair of roots, so exactly four such k are needed and no endpoint may be hit.
The answer is 2x/(x² + y²). Treat y as a constant, so the inner derivative is just 2x, and the outer logarithm contributes the reciprocal of x² + y².
For z = ln((x - sqrt(x^2 - y^2))/(x + sqrt(x^2 - y^2))) the partials are z_x = -2/sqrt(x^2 - y^2) and z_y = 2x/(y sqrt(x^2 - y^2)). Full derivation and check.
Answer: the partial with respect to a is x, and with respect to b is 1. Each time, every symbol except the one you differentiate by is frozen as a constant.
The area is π/2 + 3√3/4 ≈ 2.8698. In polar form the two circles become r = 2 sin θ and r = 4 sin θ, and the bounding lines fix θ between π/3 and π/2.
Answer: 49 pi a^4 / 32. Integrate r^3 to a^4(1-cos)^4/4, expand the fourth power, drop the odd cosine powers, and use the Wallis values pi, 3pi/4 and 5pi/8.
Answer: 4 root2/3 - pi/4 - 2/3, about 0.43355. Integrate rho^3/3, reduce sin^4 with the power-reduction formulas, then combine over theta from pi/4 to pi/2.
The answer is 1/2. The inner 1/r integration produces a logarithm of the lower limit, and the remaining theta integral reduces to the known integral of sin u ln(sin u).
The sum is -ln(1-x)/x for 0 < |x| < 1, with the removable value 1 at x = 0. Factoring out one power of x reduces it to the standard -ln(1-x) series.
The sum is the dilogarithm Li2(x), convergent for |x| <= 1. It has no elementary closed form, but gives pi^2/6 at x = 1 and -pi^2/12 at x = -1.
The substitution C(x,t) = e^(-kt) u(x,t) removes the decay term exactly, leaving the standard heat equation u_t = D u_xx with all its known solutions.
The reduction formula gives I_n = -sin^(2n-1)(x)cos(x)/(2n) + (2n-1)/(2n) I_(n-1) with I_0 = x, and the power-reduction identity gives an explicit cosine sum.
The value is (e^8 - 1)/6. Since e^(y^3) has no elementary antiderivative in y, swap to dx dy so the inner integral supplies the y^2 a substitution needs.
For a > 0 the expression equals arsinh(x/sqrt a) + (1/2)ln a. Because a is not already a square, the scale factor is sqrt a and the constant is half a log.
For a > 0 the expression equals arsinh(x/a) + ln a. Dividing inside by a matches the standard form ln(u + sqrt(u^2 + 1)), and the leftover a becomes ln a.
The limit is 0. Proof by approximating a Riemann integrable f with a step function, computing the step case exactly, and controlling the error by epsilon.
X(t) = C1 cos(t/sqrt 10) + C2 sin(t/sqrt 10). The characteristic equation r^2 + 0.1 = 0 has purely imaginary roots, so every solution oscillates forever.
Answer: (y prime)^2 = 2e^y + C1, and explicitly y = ln(C1/2) - 2 ln|sinh(sqrt(C1)(x - x0)/2)|. The trick is treating y prime as a function of y rather than x.
The solution is x = t⁴/24 + t³/3 + 8 and y = t⁴/24 − t³/3. Adding and subtracting the two equations decouples them, leaving two direct double integrations.
Answer: x = ((a/3)y + C)^3, plus the singular solution x = 0. Dividing by x^(2/3) leaves the power rule with exponent -2/3, whose antiderivative is 3x^(1/3).
The answer is y = 3/4 + Ce^(-4x). Separating gives -1/4 ln|3-4y| = x + C, and 3/4 is the stable equilibrium the solution decays toward as x grows.
Answer: 1/(2y+3) = 1/(8x+10) + C, plus the singular solution y = -3/2. Split the squared quotient so each side holds one variable, then integrate both u^-2 forms.
The answer is y = 1/(C - x). Integrating y^-2 gives -1/y = x + C, and every non-zero solution blows up at x = C after only a finite distance.
The answer is y = Ce^x. Separating gives ln|y| = x + C, and folding e^C into one constant C lets it take any real value at all, including zero.
The answer is (y - 1)^3 = 3x + C, or y = 1 + the cube root of 3x + C. Multiplying by (y-1)^2 does the separating, and the cube root needs no plus or minus.
For b > 0 the answer is y = root b tanh(root b x/a + C). Separating gives dy/(b - y^2), whose integral splits by partial fractions into a logarithm of a ratio.
The series converges: the ratio test gives L = 1/3 < 1, and the arithmetico-geometric sum formula gives the exact value 3/4. Full justification, step by step.
Answer: it diverges. The denominator has a simple zero at x = 1.9569768 inside [-pi/4, 3pi/4], so the integrand behaves like 1/(x - c), which is not integrable.
Differentiating twice gives x'' = -k^2(sin kt + cos kt), which is exactly -k^2 x. The chain rule supplies one factor of k per differentiation.
The curve falls to a minimum of −8/e² ≈ −1.0827 at x = −2, then rises to 0 at x = 0. Inflection points sit at x = −2 ± √2, and the whole graph lies at or below the axis.
The oblique asymptote at +infinity is y = 3x - 1, and the curve stays above it. Complete the square to get (3x - 1)^2 + 2, then use the conjugate trick.
The solution is y = cosh x + C. Isolating the radical and squaring cancels the p^2 terms, leaving p = (e^x - e^-x)/2 = sinh x, which integrates directly.
The general solution is y = C e^(2x) + x/2 + 1/4. The integrating factor e^(-2x) makes the left side an exact derivative, and the right side needs parts.
y = 12 + Ce^(-x/2). Rewriting it as y-prime + y/2 = 6 exposes a first-order linear ODE whose equilibrium solution y = 12 attracts every other solution.
The general solution is y = (1/5)e^(3x) + Ce^(-2x). The same integrating factor e^(2x) applies, but the right side integrates to e^(5x)/5 instead of e^x.
The general solution is y = e^(-x) + Ce^(-2x). Multiply by the integrating factor e^(2x), recognise the left side as (e^(2x)y)', integrate, then divide back through.
The general solution is y = b/a + C e^(-ax) when a is nonzero. The constant b/a is the equilibrium the solution decays towards, and a = 0 gives y = bx + C.
The general solution for t > 1 is x(t) = (t - 1)(ln(t - 1) + C). Divide by t - 1, use the integrating factor 1/(t - 1), then integrate a plain 1/(t - 1).
The implicit solution is x e^y + 3x^2 = C. The exactness test dM/dy = dN/dx = e^y passes, so a potential function exists and integrating M in x builds it.
y = (C1 + C2 x)e^(-x) + e^(3x)/16. The characteristic equation (r + 1)^2 = 0 gives the repeated root r = -1, and the trial Ae^(3x) then forces 16A = 1.
The general solution is y = C1 cos(2x) + C2 sin(2x). The characteristic equation r^2 + 4 = 0 gives r = plus or minus 2i, so the angular frequency is 2.
The general solution is y = C1 cos x + C2 sin x. The characteristic equation r^2 + 1 = 0 has roots plus or minus i, giving oscillation of angular frequency 1.
The answer is y = C1cos(x) + C2sin(x) + 12. The characteristic roots are plus and minus i, and a constant trial solution gives the particular part immediately.
The series sums to 1/2. Factoring 4n^2 - 1 into (2n-1)(2n+1) and splitting into partial fractions makes every interior term cancel in pairs.
Partial fractions give (1/2)(1/n - 1/(n+2)), a telescoping sum with a gap of two, so the first two terms survive and the total is 3/4.
The sum is e² ≈ 7.389. It is the exponential series Σxⁿ/n! evaluated at x = 2; starting the sum at n = 1 instead removes the n = 0 term and gives e² − 1.
The sum equals 4. Factoring out the 2 leaves sum k(1/2)^k, which the formula x/(1-x)^2 evaluates to 2 at x = 1/2, so the total is 2 times 2.
The answer is −ζ′(2) ≈ 0.9375482543. Differentiating ζ(s) = Σ n⁻ˢ term by term produces −ln n · n⁻ˢ, so setting s = 2 identifies the series exactly.
The sum equals ln 2 ≈ 0.6931472. Rewrite 1/(n·2ⁿ) as (1/2)ⁿ/n, match it to the Mercator series −ln(1 − x) = Σ xⁿ/n for |x| < 1, then substitute x = 1/2 and simplify.
The area is 15 root 21, about 68.74. For a plane the surface-area integrand is constant, so the answer is just root(1 + fx^2 + fy^2) times the base area.
The slope is 72 and the point of tangency is (2, 26), not (2, -2), so the tangent line is y = 72x - 118. Includes the check that fixes the printed point.
The series is (a e^-2/4b^2)(1 - x^3/12b^3 + x^4/8b^4 + ...). The x and x^2 terms cancel exactly - a substitution t = x/b makes the whole expansion parameter-free.
The degree-5 Taylor polynomial is 5x + (7/6)x^3 + (61/24)x^5. Built by multiplying the series for sin x and sec^2 x, then adding the series for sin 2x.
The series is the sum of xⁿ/n! for n ≥ 0, i.e. 1 + x + x²/2! + x³/3! + …, because every derivative of e^x is e^x and equals 1 at x = 0. It converges for all real x.
ln(sqrt(x^2+1)+x) = x - x^3/6 + 3x^5/40 - 5x^7/112 + ... on [-1,1]. The function is arcsinh x, so differentiate to 1/sqrt(1+x^2) and use the binomial series.
Answer: 1 + x - x^2/2 + x^3/2 - 5x^4/8 + ..., valid for |x| <= 1/2. Apply the binomial series with alpha = 1/2 and t = 2x, and every power of 2 is carried along.
Answer: 1 + x^2/2 - x^4/8 + x^6/16 - 5x^8/128 + ... on [-1, 1]. Substituting t = x^2 into the binomial series with alpha = 1/2 leaves only even powers of x.
The sum of 1/(sqrt(n+1)(n+2) + (n+1)sqrt(n+2)) is 1/sqrt(2). Factoring the denominator turns the term into 1/sqrt(n+1) minus 1/sqrt(n+2).
The value is pi/12. Use cylindrical coordinates around the x-axis: the surfaces meet at r = 1, the solid runs from x = r^2 up to x = r, and dV = r dx dr dtheta.
Answer: -466/5 exactly. After integrating in z up to -2x+4y-4 and in y up to 2x-2, the outer integrand is a quartic in x whose integral from 2 to 3 is -466/5.
The value is 15/2. Because the box has constant limits and the integrand is a product, integrate in x, then y, then z, pulling constants out at each stage.
The value is 19/2. Because the integrand is a product of a function of x, one of y and one of z over a box, the triple integral splits into three one-variable integrals.
Answer: 25/2 exactly. Because 5 - 5x is negative for x > 1, the y-limits run backwards; carrying the orientation through the iterated integral gives +25/2.
Answer: -188 exactly. Integrating -5y in z gives -5y(9-x^2-y^2); the y-integral collapses to -(5/4)(9-x^2)^2, and the x-integral from -1 to 1 gives -188.
Answer: 2 pi (1 - root3/2)(1 - cos 1)/3, about 0.128989. Spherical coordinates turn the cone into a fixed phi = pi/6 and make the integrand a function of rho alone.
The integral of x over the solid inside z = x² + y² up to z = 3 is 0. The region is symmetric about the plane x = 0 while the integrand is odd in x, so everything cancels.
The answer is 1/4. By symmetry each of x, y and z integrates to 1/24 over the tetrahedron, so the weighted total is just (1 + 2 + 3)/24, or 6/24.
The integral equals 336π. In cylindrical coordinates the integrand is just r², the disk radius is √(2z), and the whole computation reduces to 2π∫z² dz.
The integral equals 340π. Switch to cylindrical coordinates, use r ≤ √(2z) for 2 ≤ z ≤ 8, and split off the angular factor cos²θ sin²θ, which averages to π/4.
The value is 4*pi/5, about 2.5133. Spherical coordinates turn the integrand into rho^4 sin(phi), and the three integrals separate into 2*pi times 2 times 1/5.
The value is 1/8. Because the cube has constant limits and the integrand is a product of one-variable factors, the triple integral is just (1/2)³ — one half from each direction.
The answer is 1 − e^(−k) times the first six terms of the series for e^k. The simplex volume formula turns the triple integral into a single gamma-type integral.
The value is 0. Integrating in z then y reduces it to the integral of (x − x⁷)/6 over [−1, 1] — an odd function on a symmetric interval, so it cancels exactly.
The value is 64π/3. With z running from y² up to 4, the inner integral gives (16 − y⁴)/2, so the y-integral is 8y − y⁵/2 and the x-direction contributes a factor 2π.
The value is 32π⁵/5. The inner z-limit depends on x, not y, so integrating in z then y leaves the single power x⁴ — and the 2π upper limit produces the π⁵.
The value is 32π/3. The inner integral collapses to y⁵/2 because both the integrand and the upper z-limit involve y, and the x-direction only contributes a factor of 2π.
The integral equals pi(4 - root 2)/30, about 0.270783. Cylindrical coordinates make the cone the lower z-limit and the sphere the upper one, meeting at r = 1/root 2.
Evaluate the triple integral of 2x - 4yz over a region described by nested inequalities. The answer is -268 — full dz dy dx solution showing how each limit substitutes in.
The value is 612. Every limit runs from larger to smaller, so three sign flips multiply to one overall minus, and the innermost integration is done first.
There are two horizontal asymptotes, y = a and y = -a, whenever a is finite and nonzero. If a = 0 the two coincide and there is only the single line y = 0.
A clean bound is 1 + (pi^2/6 - 1)/ln 2 = 1.93044, from ln n >= ln 2 for n >= 2. The true sum is about 1.4976, so the bound is safe but not tight.
a = 2. Continuity at the join needs the left limit of x^2 + 2x - 3, which is 5, to equal the value 2a + 1 of the right piece, so 2a + 1 = 5.
Differentiating once gives v = 15t^2 + 3 and twice gives a = 30t, so at t = 5 seconds the velocity is 378 m/s and the acceleration is 150 m/s^2.
v(3) = 44/3 m/s. The definite integral of acceleration over [1,3] gives the change in velocity, so the initial condition at t = 1 is added at the end.
Rotating the region under y = x² + 5 from x = 0 to x = 5 about the x-axis gives V = 3500π/3 ≈ 3665.19. Full disk-method setup and evaluation, step by step.
The volume is (e^6 - 1)/2 = 201.21. See how the side of each square becomes e^x, so the volume integral is the integral of e^(2x) from 0 to 3.
The volume is πa³/3. In spherical coordinates the surface becomes ρ = a(cos φ)^(1/3), and the ρ³ that appears cancels the cube root, leaving an easy φ-integral.
The reduction formula I(n) = ((n-1)/n) I(n-2) with I(0) = pi/2 and I(1) = 1 gives the Wallis result: even n keeps a factor pi/2, odd n does not.
Almost: the correct answer is x²/2 + C. The power rule gives x²/2, but every function x²/2 + C has derivative x, so the antiderivative is a whole family, not one function.
The integral of x^7 dx is x^8/8 + C, not x^8/3 + 7. The power rule divides by the new exponent 8, and differentiating x^8/3 gives back 8x^7/3 instead of x^7.