Integration Symbol Calculator

What every part of the integral sign means, and how to evaluate one with limits
Integral from 1 to 3 of (2x + 1) dx
Integral from 0 to 2*pi of sin(x) dx
Approximate the integral from 0 to 1 of e^(-x^2) dx using Simpson's rule with n = 4
What do the numbers above and below the integral sign mean?

Anatomy of the Integral Sign

The symbol โˆซ\int is an elongated S, chosen by Leibniz for summa โ€” a sum. Every integral is the limit of a Riemann sum:

โˆซabf(x)โ€‰dx=limโกnโ†’โˆžโˆ‘i=1nf(xi)โ€‰ฮ”x\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x

Reading a definite integral left to right:

โˆซaโŸlowerย limitbโžupperย limitf(x)โŸintegranddxโŸvariable\int_{\underbrace{a}_{\text{lower limit}}}^{\overbrace{b}^{\text{upper limit}}} \underbrace{f(x)}_{\text{integrand}} \underbrace{dx}_{\text{variable}}

  • Limits aa and bb are the start and end of the interval. Without them it is an indefinite integral and the answer is a family of functions, so you must add +C+C.
  • dxdx names the variable of integration and closes the expression. It is not decoration: โˆซxโ€‰dt\int x\,dt and โˆซxโ€‰dx\int x\,dx are different problems.

Relatives: โˆฌ\iint and โˆญ\iiint integrate over a region or solid, โˆฎ\oint integrates around a closed curve, and โˆซaโˆž\int_a^\infty marks an improper integral that must be evaluated as a limit.

Evaluating an Integral With Limits

The Fundamental Theorem of Calculus turns the symbol into arithmetic. If Fโ€ฒ=fF' = f and ff is continuous on [a,b][a,b]:

โˆซabf(x)โ€‰dx=[F(x)]ab=F(b)โˆ’F(a)\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)

The square bracket with sub- and superscript is the evaluation bar โ€” it means "substitute the top, subtract the bottom".

What the result actually is. A definite integral is signed area: regions below the axis count negative. โˆซ02ฯ€sinโกxโ€‰dx=0\int_0^{2\pi}\sin x\,dx = 0 even though the curve encloses 4 units of area.

Orientation. โˆซbaf=โˆ’โˆซabf\int_b^a f = -\int_a^b f, and โˆซaaf=0\int_a^a f = 0.

When no antiderivative exists โ€” eโˆ’x2e^{-x^2} has none in elementary terms โ€” a calculator switches to numerical integration. Simpson's rule with an even nn and h=(bโˆ’a)/nh = (b-a)/n:

โˆซabfโ‰ˆh3[f0+4f1+2f2+โ‹ฏ+4fnโˆ’1+fn]\int_a^b f \approx \frac{h}{3}\Big[f_0 + 4f_1 + 2f_2 + \cdots + 4f_{n-1} + f_n\Big]

It assumes ff is smooth on [a,b][a,b]; a jump or a vertical asymptote inside the interval makes the estimate meaningless.

Common Mistakes to Avoid

  • Dropping dxdx. It marks where the integrand ends, which matters the moment you integrate a sum or substitute.
  • Forgetting +C+C on an indefinite integral, or writing +C+C on a definite one, where it cancels.
  • Subtracting in the wrong order. The evaluation bar is F(b)โˆ’F(a)F(b) - F(a), top minus bottom.
  • Calling the answer "the area". It is signed area. For total area, split at each root and integrate โˆฃfโˆฃ|f|.
  • Substituting without changing the limits. With u=g(x)u = g(x) the limits become g(a)g(a) and g(b)g(b) โ€” or convert back to xx before evaluating.
  • Running a numerical rule across a discontinuity. Simpson's and the trapezoid rule assume continuity; split the interval at the break instead.

Examples

Step 1: Find an antiderivative: F(x)=x2+xF(x) = x^2 + x
Step 2: Apply the evaluation bar: [x2+x]13\Big[x^2 + x\Big]_1^3
Step 3: Top: F(3)=9+3=12F(3) = 9 + 3 = 12
Step 4: Bottom: F(1)=1+1=2F(1) = 1 + 1 = 2
Step 5: Subtract: 12โˆ’2=1012 - 2 = 10
Answer: 1010

Step 1: F(x)=โˆ’cosโกxF(x) = -\cos x, so [โˆ’cosโกx]02ฯ€=โˆ’cosโก(2ฯ€)+cosโก(0)\Big[-\cos x\Big]_0^{2\pi} = -\cos(2\pi) + \cos(0)
Step 2: =โˆ’1+1=0= -1 + 1 = 0 โ€” the signed area cancels
Step 3: For total area, split at x=ฯ€x = \pi where sinโกx\sin x changes sign
Step 4: โˆซ0ฯ€sinโกxโ€‰dx=[โˆ’cosโกx]0ฯ€=1+1=2\int_0^{\pi}\sin x\,dx = \Big[-\cos x\Big]_0^{\pi} = 1 + 1 = 2
Step 5: โˆซฯ€2ฯ€sinโกxโ€‰dx=โˆ’1โˆ’1=โˆ’2\int_{\pi}^{2\pi}\sin x\,dx = -1 - 1 = -2, so its area is 22
Step 6: Total area =2+2=4= 2 + 2 = 4
Answer: Integral =0= 0; total enclosed area =4= 4

Step 1: eโˆ’x2e^{-x^2} has no elementary antiderivative, so use a numerical rule
Step 2: h=(1โˆ’0)/4=0.25h = (1-0)/4 = 0.25; nodes x=0,โ€‰0.25,โ€‰0.5,โ€‰0.75,โ€‰1x = 0,\, 0.25,\, 0.5,\, 0.75,\, 1
Step 3: f0=1f_0 = 1, f1=0.939413f_1 = 0.939413, f2=0.778801f_2 = 0.778801, f3=0.569783f_3 = 0.569783, f4=0.367879f_4 = 0.367879
Step 4: S=0.253[1+4(0.939413)+2(0.778801)+4(0.569783)+0.367879]S = \dfrac{0.25}{3}\big[1 + 4(0.939413) + 2(0.778801) + 4(0.569783) + 0.367879\big]
Step 5: Bracket =8.962265= 8.962265, so S=0.0833333ร—8.962265=0.746855S = 0.0833333 \times 8.962265 = 0.746855
Step 6: The true value is ฯ€2erfโก(1)=0.746824\dfrac{\sqrt{\pi}}{2}\operatorname{erf}(1) = 0.746824, so the error is about 3ร—10โˆ’53 \times 10^{-5}
Answer: โ‰ˆ0.7469\approx 0.7469

Frequently Asked Questions

They are the limits of integration: the bottom number is where the interval starts and the top number is where it ends. An integral sign with no numbers is indefinite and returns a family of antiderivatives, which is why it needs + C.

dx names the variable being integrated and marks the end of the integrand. Without it an expression like the integral of xy is ambiguous, and substitution โ€” where dx becomes du/g'(x) โ€” has nothing to transform.

It is the signed area: area above the x-axis counts positive, area below counts negative. To get the geometric area you split the interval at each root and integrate the absolute value of the function on each piece.

In LaTeX it is \int, with limits written \int_a^b. In Word, Insert > Equation gives the template, or type the Unicode character U+222B. Most solvers also accept plain text like 'integral from 0 to 1 of x^2 dx'.

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