Find the derivative of f(x) = x cubed minus 3x squared plus 2x, then evaluate its definite integral on the interval from 0 to 3.

Step-by-step solution

1. 1

   Differentiate term by term with the power rule. Each $x^n$ becomes $nx^{n-1}$ and constant multiples ride along:

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

   Nothing about the interval $[0,3]$ enters here — differentiation is a pointwise operation, so the two halves of this question are independent.

2. 2

   Antidifferentiate the same polynomial. Running the power rule backwards, $x^n\mapsto x^{n+1}/(n+1)$:

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

   The constant of integration is omitted deliberately: it cancels in the next step.

3. 3

   Apply the Fundamental Theorem of Calculus.

   $\int_0^3 f(x)\,dx=F(3)-F(0)=\left(\frac{81}{4}-27+9\right)-0=\frac{81}{4}-18=\frac{9}{4}$

4. 4

   Split at the roots to see what the $9/4$ is made of. Since $f(x)=x(x-1)(x-2)$, the graph crosses the axis at $0$, $1$ and $2$, so the integral is a signed sum:

   $\int_0^1 f=\tfrac14,\qquad \int_1^2 f=-\tfrac14,\qquad \int_2^3 f=\tfrac94$

   The first two cancel exactly, which is why the answer equals the last piece alone. Total unsigned area would instead be $\tfrac14+\tfrac14+\tfrac94=\tfrac{11}{4}$ — a different question.

5. 5

   Confirm the integral with Simpson's rule. Simpson's rule is exact for cubics, so with $a=0$, $b=3$ and midpoint $1.5$:

   $\frac{3}{6}\Bigl[f(0)+4f(1.5)+f(3)\Bigr]=\frac12\Bigl[0+4(-0.375)+6\Bigr]=\frac12(4.5)=\frac94\ \checkmark$

   The agreement is exact, not approximate, which is the strongest available check for a polynomial integrand.