Find the derivative of f(x) = x^3 - 3x^2 + 2x, then evaluate its definite integral on the interval [0, 3].

Step-by-step solution

1. 1

   Differentiate term by term with the power rule. Since $\frac{d}{dx}x^n=nx^{n-1}$ and differentiation is linear, each term can be handled on its own:

   $\frac{d}{dx}\left(x^3\right)=3x^2,\qquad \frac{d}{dx}\left(-3x^2\right)=-6x,\qquad \frac{d}{dx}(2x)=2$

   so

   $f'(x)=3x^2-6x+2$

2. 2

   Find an antiderivative for the integral. Reverse the power rule, raising each exponent by one and dividing:

   $F(x)=\frac{x^4}{4}-x^3+x^2$

   The constant of integration is omitted deliberately: in a definite integral it cancels in the subtraction, so any one antiderivative will do. Quick check: $F'(x)=x^3-3x^2+2x=f(x)$.

3. 3

   Apply the Fundamental Theorem of Calculus.

   $\int_0^3 f(x)\,dx=F(3)-F(0)$

   $F(3)=\frac{81}{4}-27+9=\frac{81}{4}-18=\frac{81-72}{4}=\frac{9}{4}$

   $F(0)=0$

   so the integral equals $\dfrac{9}{4}=2.25$.

4. 4

   Notice that this is a signed area, not a total area. Factoring, $f(x)=x(x-1)(x-2)$, so the curve crosses the axis at $x=0,1,2$: it is positive on $(0,1)$, negative on $(1,2)$, and positive again on $(2,3)$. The integral adds those pieces with sign, so it is smaller than the geometric area between the curve and the axis.

5. 5

   Verify by splitting at the roots. Using the same $F$:

   $\int_0^1 f=F(1)-F(0)=\frac14,\qquad \int_1^2 f=F(2)-F(1)=0-\frac14=-\frac14,\qquad \int_2^3 f=F(3)-F(2)=\frac94-0=\frac94$

   Adding: $\tfrac14-\tfrac14+\tfrac94=\tfrac94$, matching step 3. (The unsigned area, if it were wanted, would be $\tfrac14+\tfrac14+\tfrac94=\tfrac{11}{4}$.)