Evaluate the integral of x^2 * sin(x) from x = 0 to x = 2.

Step-by-step solution

1. 1

   Find the antiderivative by parts, twice. Taking $u=x^{2}$, $dv=\sin x\,dx$ gives $-x^{2}\cos x+\int 2x\cos x\,dx$; a second pass with $u=x$, $dv=\cos x\,dx$ gives $\int x\cos x\,dx=x\sin x+\cos x$. Together:

   $\int x^{2}\sin x\,dx=-x^{2}\cos x+2x\sin x+2\cos x+C$

   The antiderivative does not depend on the limits, so it is worth deriving once and reusing.

2. 2

   Note what changes with the limits $0$ and $2$. Here $2$ is a plain number of radians, not a multiple of $\pi$, so $\sin 2$ and $\cos 2$ do not reduce to $0$ or $\pm1$. The exact answer will therefore contain both of them, and rounding them too early is the main hazard.

3. 3

   Evaluate at the upper limit $x=2$.

   $-4\cos 2+4\sin 2+2\cos 2=4\sin 2-2\cos 2$

   The two cosine terms partially cancel: $-4\cos 2+2\cos 2=-2\cos 2$.

4. 4

   Evaluate at the lower limit $x=0$.

   $-0\cdot\cos 0+0\cdot\sin 0+2\cos 0=2$

5. 5

   Subtract to get the exact value.

   $\int_{0}^{2}x^{2}\sin x\,dx=4\sin 2-2\cos 2-2$

6. 6

   Convert to a decimal and sanity-check the sign. With $\sin 2=0.909297$ and $\cos 2=-0.416147$:

   $4(0.909297)-2(-0.416147)-2=3.637190+0.832294-2=2.469483$

   Simpson rule gives $2.4694834$ ✓. The result is positive because $\sin x>0$ on all of $[0,2]$ (since $2<\pi$).