Using Integration by Parts

Approach: The integrand is the product of $x$ and $e^x$, so take $u(x) = x$ and $v'(x) = e^x$. That shifts the derivative onto the linear factor and leaves $\int_0^1 e^x dx$, which is a basic integral.

Steps:

1. Take $u(x) = x$, $v'(x) = e^x$, so $u'(x) = 1$ and $v(x) = e^x$.
2. Apply the formula: $\int_0^1 x e^x dx = [x e^x]_0^1 - \int_0^1 e^x dx$.
3. Boundary term: $[x e^x]_0^1 = 1 \cdot e^1 - 0 \cdot e^0 = e$.
4. Remaining integral: $\int_0^1 e^x dx = [e^x]_0^1 = e - 1$.
5. Therefore $\int_0^1 x e^x dx = e - (e - 1) = 1$.

Answer: $1$

Approach: Again take $u(x) = x$, but this time $v'(x) = \cos x$, whose antiderivative is $v(x) = \sin x$. The boundary term $[x \sin x]_0^{\pi}$ happens to vanish, and the remaining $\int_0^{\pi} \sin x dx$ is a basic integral.

Steps:

1. Take $u(x) = x$, $v'(x) = \cos x$, so $u'(x) = 1$ and $v(x) = \sin x$.
2. Apply the formula: $\int_0^{\pi} x \cos x dx = [x \sin x]_0^{\pi} - \int_0^{\pi} \sin x dx$.
3. Boundary term: $[x \sin x]_0^{\pi} = \pi \sin \pi - 0 \sin 0 = 0$.
4. Remaining integral: $\int_0^{\pi} \sin x dx = [-\cos x]_0^{\pi} = -\cos \pi + \cos 0 = 1 + 1 = 2$.
5. Therefore $\int_0^{\pi} x \cos x dx = 0 - 2 = -2$.

Answer: $-2$

Approach: One application of the formula leaves $\int_0^1 2x e^x dx$, which is still a product, so the formula has to be used twice: the first round removes $x^2$, the second one removes $x$.

Steps:

1. First round: take $u(x) = x^2$, $v'(x) = e^x$, so $u'(x) = 2x$ and $v(x) = e^x$.
2. $\int_0^1 x^2 e^x dx = [x^2 e^x]_0^1 - \int_0^1 2x e^x dx = e - 2\int_0^1 x e^x dx$.
3. Integrate $\int_0^1 x e^x dx$ by parts once more (take $u(x) = x$, $v'(x) = e^x$): $\int_0^1 x e^x dx = [x e^x]_0^1 - \int_0^1 e^x dx = e - (e - 1) = 1$.
4. Substitute back: $\int_0^1 x^2 e^x dx = e - 2 \cdot 1 = e - 2$.

Answer: $e - 2$