Problems

Solution:

1. Let \(2t = u\), \begin{align*} \int_2^x \sqrt{\frac{u-2}{u+2}} du &= \int_{t=1}^{t=x/2} \sqrt{\frac{2t-2}{2t+2}}2 \d t \\ &= 2\int_{t=1}^{x/2} \sqrt{\frac{t-1}{t+1}} \d t \\ &= 2f\l\frac{x}{2}\r \end{align*}
2. Let \(v = u-2\), \begin{align*} \int_0^x \sqrt{\frac{u}{u+4}} du &= \int_{v = 2}^{x+2} \sqrt{\frac{v-2}{v+2}} \d v \\ &= 2 f \l \frac{x+2}{2} \r \end{align*}
3. Let \(v = u-2, \d v = \d u\) \begin{align*} \int_5^x \frac{u-5}{u+1} du &= \int_3^{x-2} \frac{v-3}{v+3} \d v \\ &= \int_1^{\frac{x-2}{3}} \frac{3t - 3}{3t+3} 3 \d t \\ &= 3 f \l \frac{x-2}{3} \r \end{align*}
4. Let \(v = u^2, \d v = 2u \d u\)\begin{align*}\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du &= \int_1^2 \sqrt{\frac{u^2}{u^2+4}} u \d u \\ &= \int_1^4 \sqrt{\frac{v}{v+4}} \frac12 \d v \\ &= f \l \frac{4+2}{2} \r - f \l \frac{3}{2} \r \\ &= f(3) - f(\frac32) \end{align*}

Solution:

1. Claim: If \(f(x)\) non-zero for some \(x \in [0,1]\) and \(\int_0^1 f(x) \d x =0\) then \(f\) takes both positive and negative values in the interval \([0,1]\). Proof: Suppose not, then WLOG suppose \(f(x) > 0\) for some \(x \in [0,1]\). Then notice (since \(f\) is continuous) that there is some interval where \(f(x) > 0\) around the \(x\) we have already shown exists. But then \(\int_{\text{interval}} f(x) \d x > 0\) and since \(f(x) \geq 0\) everywhere \(\int_0^1 f(x) \d x > 0\), which is a contradiction.
2. \(\,\) \begin{align*} && \int_0^1 (x - \alpha)^2 g(x) \d x &= \int_0^1 x^2g(x) \d x - 2 \alpha \int_0^1 x g(x) \d x + \alpha^2 \int_0^1 g(x) \d x \\ &&&= \alpha^2 - 2\alpha \cdot \alpha + \alpha^2 \cdot 1 \\ &&&= 0 \end{align*} Therefore \(g(x)(x-\alpha)^2\) is a continuous function which is either exactly \(0\) (in which case we've already found our \(0\)) or it is a continuous function which is both positive somewhere and has \(0\) integral, and therefore by part (i) must take both positive and negative values (and therefore takes \(0\) in between those points by continuity). \begin{align*} &&1 &= \int_0^1 a+bx \d x \\ &&&= a + \frac12 b \\ && \alpha &= \int_0^1 ax + bx^2 \d x \\ &&&= \frac12 a + \frac13 b \\ && \alpha^2 &= \int_0^1 ax^2 + bx^3 \d x\\ &&&= \frac13 a + \frac14 b \\ \Rightarrow && \frac1{36}(3a+2b)^2 &= \frac1{12}(4a+3b) \\ \Rightarrow && \frac1{36}(3a+2(2-2a))^2 &= \frac1{12}(4a+3(2-2a)) \\ \Rightarrow && (4-a)^2 &= 3(6-2a) \\ \Rightarrow && 16-8a+a^2 &= 18-6a \\ \Rightarrow && a^2-2a-2 &= 0 \\ \Rightarrow && (a-1)^2 - 3 &= 0 \\ \Rightarrow && a &= \pm \sqrt{3}+1 \\ && b &= \mp 2\sqrt{3} \end{align*} So say \(a = \sqrt{3}+1, b = -2\sqrt{3}\) This has a root at \(-\frac{a}{b} = \frac{1+\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3}+3}{6} < 1\) so we have met our condition.
3. Consider \(h'\), we must have \begin{align*} && \int_0^1 h'(x)\d x &= h(1)-h(0) =1\\ && \int_0^1 xh'(x) \d x &= \left [x h(x) \right]_0^1 - \int_0^1 h(x) \d x \\ &&&= 1 - \beta \\ && \int_0^1 x^2 h'(x) \d x &= \left [ x^2h(x) \right]_0^1 - \int_0^1 2xh(x) \d x \\ &&&= 1 - 2\tfrac12 \beta(2-\beta) \\ &&&= (1-\beta)^2 \end{align*} Therefore \(h'\) satisfies the conditions with \(\alpha = 1-\beta\), so \(h'\) must have a root in our interval.

Solution:

1. Let \(f(x) = 1, g(x) = e^x, a = 0, b = t\), so \begin{align*} && \left ( \int_0^t e^x \d x \right)^2 &\leq \left (\int_0^t 1^2 \d x \right) \cdot \left (\int_0^t (e^x)^2 \d x \right) \\ \Rightarrow && (e^t-1)^2 &\leq t \cdot (\frac12e^{2t} - \frac12) \\ \Rightarrow && \frac{e^t-1}{e^t+1} & \leq \frac{t}{2} \end{align*}
2. Let \(f(x) = x, g(x) = e^{-\frac14 x^2}, a = 0, b = 1\) \begin{align*} && \left ( \int_0^1 xe^{-\frac14 x^2} \d x \right)^2 &\leq \left (\int_0^1 x^2 \d x \right) \cdot \left (\int_0^1 (e^{-\frac14x^2})^2 \d x \right) \\ \Rightarrow && \left ( \left [-2e^{-\frac14x^2} \right]_0^1 \right)^2 & \leq \frac{1}{3} \int_0^1 e^{-\frac12 x^2} \d x \\ \Rightarrow && \int_0^1 e^{-\frac12 x^2} \d x & \geq 12(1-e^{-\frac14})^2 \end{align*}
3. Let \(f(x) = 1, g(x) = \sqrt{\sin x}, a = 0, b = \tfrac12 \pi\), then \begin{align*} && \left ( \int_0^{\frac12 \pi} \sqrt{\sin x} \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} 1^2 \d x \right) \cdot \left (\int_0^{\frac12 \pi}|\sin x| \d x \right) \\ &&&= \frac{\pi}{2} \cdot 1 \\ \Rightarrow && \int_0^{\frac12 \pi} \sqrt{\sin x} \d x & \leq \sqrt{\frac{\pi}{2}} \end{align*} Let \(f(x) =(\sin x)^{\frac14}, g(x) = \cos x, a = 0, b = \tfrac12 \pi\), so \begin{align*} && \left ( \int_0^{\frac12 \pi} (\sin x)^{\frac14} \cos x \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} \cos^2 x \d x \right) \cdot \left (\int_0^{\frac12 \pi}\sqrt{\sin x} \d x \right) \\ \Rightarrow &&\left ( \left [\frac45 (\sin x)^{\frac54} \right]_0^{\frac12 \pi} \right)^2 & \leq \frac{\pi}{4} \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \\ \Rightarrow && \frac{64}{25\pi} &\leq \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \end{align*}

Solution:

1. \(\,\) \begin{align*} && \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\ \Rightarrow && b &= \frac{4}{3} \end{align*}
2. \(\,\) \begin{align*} && \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\ \Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\ \Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\ \Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\ &&&= (b-1)(3b^3-b^2-7b-7) \\ \Rightarrow && 0 &= 3b^3-b^2-7b-7 \end{align*}
   

   + - ↺
   Let \(f(x) = 3x^3-x^2-7x-7\) then \(f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0\), \(f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0\) therefore the root must lie between \(2\) and \(3\).
3. \(,\) \begin{align*} && \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\ \Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\ \Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\ \Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\ \Rightarrow && 3p^2 + q^2 &= 3qp^2 \\ \Rightarrow && 3p^2(q-1) &= q^2 \\ \Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\ \Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\ \Rightarrow && 3(q-1) &\leq 1 \\ \Rightarrow && q & \leq \frac{4}{3} \\ \end{align*}

Solution: \begin{align*} && T &= \int_{\frac13}^{\frac12} \frac{{\rm artanh}\, t}t \,\d t \\ && &=\int_{\frac13}^{\frac12} \frac{1}{2t}\ln \left ( \frac{1+t}{1-t} \right) \,\d t \\ u = \tfrac{1+t}{1-t}, t= \tfrac{u-1}{u+1}, \d t = \tfrac{2}{(u+1)^2} \d t &&&= \int_{u=2}^{u=3} \frac{1}{2t} \ln u \frac{2}{(u+1)^2} \d u \\ &&&= \int_2^3 \frac{u+1}{u-1} \ln u \frac{1}{(u+1)^2} \d u \\ &&&= \int_2^3 \frac{1}{u^2-1} \ln u \d u \end{align*} \begin{align*} && U&= \int _{\ln 2 }^{\ln 3 } \frac{u}{2\sinh u}\, \d u \\ v = e^u, \d v = e^u \d u &&&= \int_{v=2}^{v=3} \frac{\ln v}{v - \frac{1}{v}} \frac{1}{v} \d v \\ &&&= \int_2^3 \frac{1}{v^2-1} \ln v \d v \end{align*} \begin{align*} &&V &= - \int_{\frac13}^{\frac12} \frac{\ln v}{1-v^2} \,\d v \\ u = \tfrac1v, \d u = -\tfrac1{v^2} \d v &&&= -\int_{u=3}^{u=2} \frac{-\ln u}{1 - \frac{1}{u^2}} \frac{-1}{u^2} \d u \\ &&&= -\int_3^2 \frac{\ln u}{u^2-1} \d u \\ &&&= \int_2^3 \frac{1}{u^2-1} \ln u \d u \end{align*} \begin{align*} &&X&= \int _{\frac12\ln2}^{\frac12\ln3} \ln ({\coth x})\, \d x \\ u = \coth x, \d u =(1-u^2) \d x &&&= \int_{u = 3}^{u=2} \ln u \frac{1}{1-u^2} \d u \\ &&&= \int_2^3 \frac{\ln u}{u^2-1} \d u \end{align*} Therefore all integrals are equal to the same integral, namely \(\displaystyle \int_2^3 \frac{\ln u}{u^2-1} \d u\)

Solution:

1. \begin{align*} && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ u = v^{-1}, \d u = -v^{-2} \d v&&&= \int_{v = \infty}^{v = 1/x} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= \int_{1/x}^\infty \frac{1}{1+v^2} \d v \\ \\ \Rightarrow && t(x) + t(1/x) &= \int_0^x \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_{1/x}^{\infty} \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u \\ &&&= \frac12 p \\ \\ \Rightarrow && t(1) +t(1/1) = 2t(1) &= \frac12 p \end{align*}
2. \(\,\) \begin{align*} && y &= \frac{u}{\sqrt{1+u^2}} \\ \Rightarrow && y^2 &= \frac{u^2}{1+u^2} \\ &&&= 1-\frac{1}{1+u^2} \\ \Rightarrow && 1+u^2 &= \frac{1}{1-y^2} \\ \Rightarrow && u &= \frac{y}{\sqrt{1-y^2}} \\ \\ && \frac{\d u}{\d y} &= \frac{\sqrt{1-y^2} + y^2(1-y^2)^{-1/2}}{1-y^2} \\ &&&= \frac{1}{(1-y^2)^{3/2}} \\ \\ && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_0^{y = x/\sqrt{1+x^2}} \frac{1}{1 + \frac{y^2}{1-y^2}} \frac{1}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1-y^2}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1}{(1-y^2)^{1/2}} \d y \\ &&&= s\left ( \frac{x}{\sqrt{1+x^2}} \right) \\ \\ \Rightarrow && s\left ( \frac{1}{\sqrt{2}} \right) &= t(1) = \frac14p \end{align*}
3. \(\,\) \begin{align*} && z &= \frac{u + \frac1{\sqrt{3}}}{1- \frac{1}{\sqrt{3}} u}\\ \Rightarrow && z - \frac{z}{\sqrt{3}}u &= u + \frac{1}{\sqrt{3}} \\ \Rightarrow && u &= \frac{z-\frac{1}{\sqrt{3}}}{1 + \frac{z}{\sqrt{3}}} \\ \\ \Rightarrow && \frac{\d u}{\d z} &= \frac{\sqrt{3}(\sqrt{3}+z ) -(\sqrt{3}z-1)}{\left (\sqrt{3}+z \right)^2} \\ &&&= \frac{4}{(\sqrt{3}+z)^2} \\ \\ \Rightarrow && t \left ( \frac{1}{\sqrt{3}} \right) &= \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \d u \\ &&&= \int_{z=1/\sqrt{3}}^{z=\sqrt{3}} \frac{1}{1 + \left ( \frac{\sqrt{3}z-1}{\sqrt{3}+z}\right)^2} \frac{4}{(\sqrt{3}+z)^2} \d z\\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{(\sqrt{3}+z)^2+(\sqrt{3}z-1)^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{4+4z^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \d z \end{align*} Notice that \(t(1/\sqrt{3})+t(\sqrt{3}) = \frac12p\) and also notice that \(t(1/\sqrt{3}) + t(1/\sqrt{3}) =t(\sqrt{3})\) so \(3t(1/\sqrt{3}) = \frac12p\)

Solution:

1. \begin{align*} \int_{-\pi}^\pi |\sin x | \d x &= \int_{-\pi}^{0} - \sin x \d x + \int_0^\pi \sin x \d x \\ &= \left [\cos x \right]_{-\pi}^{0} +[-\cos x]_0^{\pi} \\ &= 1-(-1)+(1)-(-1) \\ &= 4 \end{align*}
2. \begin{align*} \int_{-\pi}^\pi \sin | x | \d x &= \int_{-\pi}^0 - \sin x \d x + \int_0^\pi \sin x \d x \\ &= 4 \end{align*}
3. \begin{align*} \int_{-\pi}^\pi x \sin x \d x &= \left [ -x \cos x \right]_{-\pi}^\pi + \int_{-\pi}^{\pi} \cos x \d x \\ &= \pi -(-\pi) + \left [\sin x \right]_{-\pi}^\pi \\ &= 2\pi \end{align*}
4. \begin{align*} \int_{-\pi}^{\pi} x^{10} \sin x \d x &\underbrace{=}_{x^{10}\sin x \text{ is odd}} 0 \end{align*}

Solution:

1. \begin{align*} u = \frac{\pi}{2} - x :&& \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_{\frac12\pi}^0 \ln (\cos u) (- 1)\d u \\ &&&= \int_0^{\frac12 \pi} \ln (\cos x) \d x \\ \Rightarrow && 2 \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_0^{\tfrac12 \pi} \ln (\sin x) \d x +\int_0^{\tfrac12 \pi} \ln (\cos x) \d x \\ &&&= \int_0^{\tfrac12 \pi}\left (\ln (\sin x)+ \ln (\cos x) \right) \d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\frac12 \sin 2x \right) \d x \\ &&&= \int_0^{\frac12 \pi} \left ( \ln \left (\sin 2x \right) - \ln 2 \right)\d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{2} \ln 2\\ \Rightarrow && \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \frac12 \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{4} \ln 2 \end{align*} \begin{align*} u = 2x, \d u = 2 \d x && \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x &= \int_0^{\pi} \ln (\sin u) \frac12 \d u \\ &&&= \frac12 \int_0^{\pi} \ln (\sin u) \d u \\ &&&=\frac12 \left ( \int_0^{\pi/2} \ln (\sin u) \d u + \int_{\pi/2}^{\pi} \ln (\sin u) \d u \right)\\ &&&= \int_0^{\pi/2} \ln (\sin u) \d u \\ \Rightarrow && I &= \frac12 I - \frac14 \pi \ln 2 \\ \Rightarrow && I &= -\frac12 \pi \ln 2 \end{align*}
2. \begin{align*} && \ln u &< u & \quad (u \geq 1)\\ \underbrace{\Rightarrow}_{u = \sqrt{x}} && \ln \sqrt{x} &< \sqrt{x} \\ \Rightarrow && \frac12 \ln x &< \sqrt{x} \\ \Rightarrow && \frac{\ln x}{x} &< \frac{2\sqrt{x}}{x} \\ &&&= \frac{2}{\sqrt{x}} \\ &&&\to 0 & (x \to \infty) \\ && x \ln x &= \frac{\ln 1/y}{y} \\ &&&= -\frac{\ln y}{y} \\ &&&\to 0 & (y \to \infty, x \to 0) \end{align*}
3. \begin{align*} \int_{0}^{\frac{1}{2}\pi}x\cot x\,\mathrm{d}x &= \left [ x \ln(\sin x) \right]_0^{\pi/2} - \int_0^{\pi/2} \ln (\sin x) \d x \\ &= \left ( \frac{\pi}{2} \ln 1 - \lim_{x \to 0} x \ln (\sin x) \right) - \left ( -\frac12 \pi \ln 2 \right) \\ &= \frac12 \pi \ln 2 \end{align*}