====== HW 15 ======
This is for your practice only, not going to be graded. The solution will be release next Wednesday.

=== 1. Ross Ex 33.9 ===
(uniform convergence and integral). 

See Rudin Thm 7.16 (page 151)

=== 2. Ross Ex 33.5 ===
(bound an integral by replacing an part of the integrand by something nice)

We want to show
$$| \int_{-2\pi}^{2 \pi} x^2 \sin^8(e^x) dx |  \leq 16 \pi^3 /3 $$
since $|\sin^8(e^x)| \leq 1$, we have
$$| \int_{-2\pi}^{2 \pi} x^2 \sin^8(e^x) dx | \leq   \int_{-2\pi}^{2 \pi} |x^2 \sin^8(e^x)| dx  \leq \int_{-2\pi}^{2 \pi} |x^2| dx = x^3/3|^{2\pi}_{-2\pi} =  16 \pi^3 /3 $$


=== 3. Ross Ex 33.14 (a). ===
For any continuous $g(x) \geq 0$ on $[a,b]$ and continuous $f(x)$, we want to show
$$ \int_a^b f(t) g(t) dt = f(x)  \int_a^b  g(t) dt $$
for some $x \in [a,b]$. 

Proof: If $g(t) =0 $ for all $t \in [a,b]$ then both sides are zero, and there is nothing to prove. Otherwise, assume $g(t) \neq 0$ for some $t \in [a,b]$. By continuity, $g(t)$ is non-zero hence positive on an open subset, hence  $\int_a^b g(t) dt > 0$. 

let $M = \sup \{f(t): t \in [a,b]\}$ and $ m= \inf \{f(t): t \in [a,b]\}$, then since $g(t) \geq 0$, we have $m g(t) \leq f(t) g(t) \leq M g(t)$, thus 
$$ m \int_a^b  g(t) dt \leq \int_a^b f(t) g(t) dt \leq M \int_a^b  g(t) dt. $$
Thus, $$ A = \frac{\int_a^b f(t) g(t) dt}{ \int_a^b  g(t) dt} \in [m, M]$$
by intermediate value theorem, there is $x \in [a,b]$, such that $f(x) = A$. This finishes the proof. 


=== note===
For the following two problems, if $a > b$, then $\int_a^b f(t) dt = - \int_b^a f(t) dt $. 

=== 4. Ross Ex 34.5,===


Let $f$ be continuous on $\R$, and let $F(x)$ be defined by
$$ F(x) = \int_{x-1}^{x+1} f(t) dt $$
show that $F$ is differentiable on $\R$ and compute $F'(x)$. 

Solution: we prove by definition. For any $\epsilon \neq 0$, we have 
$$ F(x+\epsilon) - F(x) = \int_{x+\epsilon-1}^{x+\epsilon+1} f(t) dt - \int_{x-1}^{x+1} f(t) dt
= \int_{x+1}^{x+\epsilon+1} f(t) dt - \int_{x-1}^{x+\epsilon-1} f(t) dt $$
Divide by $\epsilon$ and taking limit, we have
$$ \lim_{\epsilon \to 0} \frac{F(x+\epsilon) - F(x)}{\epsilon} = \lim_{\epsilon \to 0^+} \epsilon^{-1} \int_{x+1}^{x+\epsilon+1} f(t) dt - \epsilon^{-1} \int_{x-1}^{x+\epsilon-1} f(t) dt = f(x+1) - f(x-1) $$
where we used fundamental theorem of calculus. 

Hence $$F'(x) = f(x+1) - f(x-1) $$

=== 34.6. ===
Let $f$ be continuous function on $\R$, let 
$$ G(x) = \int^{sin(x)}_0 f(t) dt $$
prove that $G(x)$ is differentiable, and compute $G'(x)$.

Define $H(u) = \int^{u}_0 f(t) dt $, for any $u \in \R$, then $G(x) = H(\sin(x))$. Since $H(u)$ is differentiable, with $H'(u) = f(u)$, and $\sin(x)$ is differentiable, we have $G(x)$ is differentiable, with 
$$ G'(x) = H'(\sin(x)) \cos(x) = f(\sin(x)) \cos(x) $$

=== 6, 35.3 ===
See definition 35.2 for Stieljes integral. In particular, if $F$ has a jump on the integration domain's boundary, those points are considered in the integral. Hence
$$ \int_a^b f(t) d F(t) = \sum_{n \in [a,b] \cap \Z} f(n) $$
For example, 
$$ \int_0^6 x dF(x) = 0 + 1 + \cdots + 6 = .. $$

=== 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
$$ \int_0^{\pi/2} x d F(x) = x \sin(x)|^{\pi/2}_0 - \int_0^{\pi/2} \sin(x) dx = \pi/2 - (-\cos(x))|^{\pi/2}_0 = \pi/2 -1. $$