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

(uniform convergence and integral).

See Rudin Thm 7.16 (page 151)

(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 $| \int_{-2\pi}^{2 \pi} x^2 \sin^8(e^x) dx | \leq \int_{-2\pi}^{2 \pi} |x^2 \sin^8(e^x)| dx \leq \int_{-2\pi}^{2 \pi} |x^2| dx = x^3/3|^{2\pi}_{-2\pi} = 16 \pi^3 /3$

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.

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

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

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 $G'(x) = H'(\sin(x)) \cos(x) = f(\sin(x)) \cos(x)$

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, $\int_0^6 x dF(x) = 0 + 1 + \cdots + 6 = ..$

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