\int_1^\infty f(x)\,dx =\lim_{b\to \infty} \int_1^bf(x)\,dx,
\] and the latter expression $ \int_1^bf(x)\,dx$ is an increasing function in $b$, hence $ \int_1^bf(x)\,dx$ either converges or diverge to $\infty$. And in this case,
$\dis \int_1^\infty f(x)\,dx$ exists $\iff$ $\dis \int_1^\infty f(x)\,dx$ is bounded.
Problem 1. (a) F; (b) F; (c) F; (d) F; (e) T; (f) F
Problem 2. $\dis \int_0^{1/4}\frac{1}{\sqrt{x}(1-x)}\,dx$ converges since the limit \[
\lim_{\epsilon \to 0^+}\int_\epsilon^{1/4} \frac{dx}{\sqrt{x}(1-x)} =\lim_{\epsilon\to 0^+} \brac{\ln 3-\ln \frac{1+\sqrt{\epsilon}}{1-\sqrt{\epsilon}}}
\] exists and is equal to $\ln 3$.
Next $\dis \int_0^1\frac{1}{\sqrt{x}(1-x)}\,dx$ diverges since \[
\int_0^1\frac{1}{\sqrt{x}(1-x)}\,dx\ge \int_0^1\frac{1}{1-x}\,dx=\infty.
\]
Problem 3. Since \[
\frac{1}{\sqrt[3]{x^4+1}}\leq \frac{1}{x^{4/3}}\quad\text{and}\quad \frac{1}{x-\sin x}\ge \frac{1}{x},
\] (a) converges and (b) diverges. For (c), since \[
\int_{-1}^1\frac{1}{x}\,dx \mathop{=}\limits^{\mathrm{def}}\lim_{a\to 0^-}\int_{-1}^a\frac{1}{x}\,dx+\lim_{b\to 0^+}\int_b^1 \frac{1}{x}\,dx,
\] since both of them diverge, the integral $\int_{-1}^1\frac{1}{x}\,dx$ must also diverge.
Problem 4. By integration by parts one has \[
\int_0^a\frac{x\tan^{-1}(x)}{(1+x^2)^2}\,dx = -\frac{1}{2}\brac{\frac{\tan^{-1}a}{1+a^2}-\int_0^a \frac{1}{(1+x^2)^2}\,dx},
\] since the latter integral can be computed by the substitution $x=\tan\theta$, we have \[
\int_0^a\frac{x\tan^{-1}(x)}{(1+x^2)^2}\,dx=-\frac{1}{2}\brac{\frac{\tan^{-1}a}{1+a^2}-\frac{\tan^{-1} a+\frac{1}{2}\sin (2\tan^{-1}a)}{2} },
\] hence the integral converges to $\dis \frac{\pi}{8}$.
Problem 5. Denote $\dis I=\int_0^\pi \ln (\sin x)\,dx$. To determine convergence, let's consider $\dis \int_\epsilon^\pi$ first. By integration by parts, we have \begin{equation}\label{existence}
\int_\epsilon^{\pi/2} \ln(\sin x)\,dx = -\epsilon\ln( \sin\epsilon) -\int_\epsilon^{\pi/2}\frac{x\cos x}{\sin x}\,dx.
\end{equation} The first term go to zero because \[
\epsilon\ln( \sin\epsilon) = \epsilon \ln \epsilon +\epsilon \ln \brac{\frac{\sin \epsilon}{\epsilon}}\to 0 + 0\times 0 = 0.
\] For the second term, since $\dis \frac{x\cos x}{\sin x} \ge 0$ on $(0,\pi/2]$. We can try to use comparison test. Since $\dis \sin x\ge \frac{2x}{\pi}$ on $[0,\pi/2]$, this is derived from the comparison of the graph of $y=\sin x$ and the straight line joining $(0,0)$ and $(\pi/2,1)$, we conclude \[
\int_\epsilon^{\pi/2}\frac{x\cos x}{\sin x}\,dx \leq \frac{\pi}{2}\int_\epsilon^{\pi/2} \cos x\,dx\leq \frac{\pi}{2}\brac{\frac{\pi}{2}-\epsilon}.
\] Hence if we take $\epsilon\to 0$, $\dis \int_0^{\pi/2}\frac{x\cos x}{\sin x}\,dx$ converges. By (\ref{existence}), $I$ exists.
Now we try to compute the integral. A change of variable shows that \[
\int_{\pi/2}^{\pi} \ln (\sin x)\,dx = \int_{0}^{\pi/2} \ln (\cos x)\,dx,
\] hence \[\begin{align*}
I&=\int_{0}^{\pi/2} \ln (\sin x)\,dx + \int_{\pi/2}^{\pi} \ln (\sin x)\,dx\\
& = \int_0^{\pi/2}\ln (\sin x\cos x)\,dx\\
&=\int_0^{\pi/2}\ln \brac{\frac{\sin 2x}{2}}\,dx\\
&=\int_0^{\pi/2}(\ln \sin 2x - \ln 2)\,dx\\
&=\int_0^{\pi/2}\ln \sin 2x\,dx - \frac{\pi}{2}\ln 2\\
&=\frac{1}{2}I -\frac{\pi \ln 2}{2}\\
I&=-\pi\ln 2.
\end{align*}
\]
Problem 6. (a) Let's choose $0<s<t$, then as $f$ is decreasing, we have \[
\int_s^tf(x)\,dx \ge \int_s^t f(t)\,dx = (t-s)f(t),
\] since $\int_0^\infty f\,dx$ exists, $\int_s^tf(x)\,dx$ will go to zero when both $s$ and $t$ are large. We will elaborate later. Now \[
tf(t)\leq \frac{t}{t-s}\int_s^tf(x)\,dx,
\] we try to take $s=t/2$, then we have \[
tf(t)\leq \frac{t}{\frac{t}{2}}\int_{t/2}^{t}f(x)\,dx=2\brac{\int_0^t f(x)\,dx - \int_0^{t/2}f(x)\,dx},
\] since \[
\lim_{t\to\infty}2\brac{\int_0^t f(x)\,dx - \int_0^{t/2}f(x)\,dx} = 2\brac{\int_0^\infty f\,dx - \int_0^\infty f\,dx}=0,
\] by Sandwich theorem we conclude $\lim_{t\to\infty}tf(t)=0$.
(b) Firstly let's observe that \[
\frac{1}{n}\int_0^nxf(x)\,dx =\int_0^1 nx f(nx)\,dx.
\] Now as in part (a) we try to consider \[\begin{align}
\nonumber \int_{\sqrt{n}}^n f(x)\,dx &=\int_{1/\sqrt{n}}^1nf(nx)\,dx\\
\nonumber& \ge \int_{1/\sqrt{n}}^1 nx f(nx)\,dx\\
\nonumber&=\int_0^1nx f(nx) \,dx - \int_0^{1/\sqrt{n}}nx f(nx)\,dx\\
\label{here}&=\frac{1}{n}\int_0^nxf(x)\,dx - \underbrace{\int_0^{1/\sqrt{n}}nx f(nx)\,dx}_{I_n}.
\end{align}
\] Now we are almost done since $\limn \int_{\sqrt{n}}^n f(x)\,dx =0$, we hope that $\limn I_n=0$, but this is simple since \[
I_n\leq \int_0^{1/\sqrt{n}}\sqrt{n}f(nx)\,dx =\frac{1}{\sqrt{n}}\int_0^{\sqrt{n}}f(x)\,dx\leq \frac{1}{\sqrt{n}}\int_0^\infty f(x)\,dx,
\] continuing from (\ref{here}) we have \[
\frac{1}{n}\int_0^nxf(x)\,dx \leq I_n + \int_{\sqrt{n}}^nf(x)\,dx\leq \frac{1}{\sqrt{n}}\int_0^\infty f(x)\,dx+\int_{\sqrt{n}}^nf(x)\,dx,
\] by Sandwich theorem, $\limn \frac{1}{n}\int_0^nxf(x)\,dx=0$.
Problem 7. (a) We can rewrite the equation as follows \[
e^{2y}\,dy =e^x\,dx,
\] we then integrate both sides to get \[
e^{2y}/2=e^x+C\implies e^{2y}=2e^x+2C \implies 2y = \ln(2C+2e^x),
\] so there is a constant $D$ such that \[
y=\frac{1}{2}\ln (D+2e^x).
\]
(b) We rewrite the equation as follows \[
e^{-y}\,dy = e^{\sin x}\cos x\,dx\implies \frac{e^{-y}}{-1} = e^{\sin x}+C\implies e^{-y}=-e^{\sin x}-C,
\] hence there is a constant $D$ such that \[
y=\ln \frac{1}{D-e^{\sin x}}.
\]
Problem 8. (a) The equation can be written as \[
\frac{dh}{dt}=-0.2\sqrt{h}=\implies \int h^{-1/2}\,dh =-0.2 \int \,dt\implies h^{1/2}=-\frac{1}{10}t+\frac{C}{2},
\] for some constant $C$, write $D=C/2$, then \[
h=\sqrt{D-\frac{t}{10}},
\] for some constant $D$. By the initial condition we have $h(0) = 0.5$, hence \[
\frac{1}{2}=\sqrt{D}\implies D = \frac{1}{4},
\] and we have \[
h(t) = \sqrt{\frac{1}{4}-\frac{t}{10}}.
\] (b) When $h(t)=0$, $\frac{1}{4}-\frac{t}{10}=0$, hence $t=\frac{5}{2}$.
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