问题 1.
Evaluate the integral
$$
\int_{0}^{\infty} x e^{-x} d x
$$
Solution: The exponential can be expressed as a derivative:
$$
e^{-x}=\frac{d}{d x}\left(-e^{-x}\right)
$$
证明 .
in this case, then, $f(x)=x, g(x)=-e^{-x}$, and $d f / d x=1$, so
$$
\int_{0}^{\infty} x e^{-x} d x=\int_{0}^{\infty} e^{-x} d x-\left.x e^{-x}\right|{0} ^{\infty}=-\left.e^{-x}\right|{0} ^{\infty}=1
$$
PHAS0038 COURSE NOTES ：
(a) $\int_{2}^{6}\left(3 x^{2}-2 x-1\right) \delta(x-3) d x$.
(b) $\int_{0}^{5} \cos x \delta(x-\pi) d x$.
(c) $\int_{0}^{3} x^{3} \delta(x+1) d x$.
(d) $\int_{-\infty}^{\infty} \ln (x+3) \delta(x+2) d x$.