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--- math104-f21:hw15 [2021/12/09 00:49]
pzhou
+++ math104-f21:hw15 [2021/12/09 00:54]
pzhou
@@ Line 62: / Line 62: @@
 $$ \int_0^6 x dF(x) = 0 + 1 + \cdots + 6 = .. $$
-, 35.4
+=== 7, 35.4 ===
+Since $F(t)$ is differentiable and monotone over that range, we have $dF(x) = F'(x) dx$, with $F'(x) = \cos(x)$ for $t \in [-\pi/2, \pi/2]$.
+$$ \int_0^{\pi/2} x d F(x) = \int_0^{\pi/2} x \cos(x) d x $$
+Alternatively, one can compute using integration by part. If $f$ is also monotone and differentiable, then
+$$ \int_a^b f dF = \int_a^b d(f F) - F df = f F|^b_a - \int_a^b F df(x) $$
+Here, in this problem, we have $f(x) = x$, thus
+$$ \int_0^{\pi/2} x d F(x) = x \sin(x)|^{\pi/2}_0 - \int_0^{\pi/2} \sin(x) dx = \pi/2 - (-\cos(x))|^{\pi/2}_0 = \pi/2 -1. $$

Lecture Notes