Problems

Solution:

1. \(\,\) \begin{align*} (x \geq 1): && \int_1^x \frac{1}{t} \d t &\leq \int_1^x 1 \d t \\ \Rightarrow && \ln x - \ln 1 &\leq x - 1 \\ \Rightarrow && \ln x & \leq x - 1 \\ \\ (0 < x \leq 1):&& \int_x^1 1\d t &\leq \int_x^1 \frac{1}{t} \d t \\ \Rightarrow&& 1- x &\leq \ln 1 - \ln x \\ \Rightarrow&& \ln x &\leq x - 1 \end{align*}
2. \(\,\) \begin{align*} (x \geq 1): && \int_1^x \frac{1}{t^2} \d t &\leq \int_1^x \frac{1}{t} \d t \\ \Rightarrow && -\frac1x+1 &\leq \ln x - \ln 1 \\ \Rightarrow && 1 - \frac1x &\leq \ln x \\ \\ (0 < x \leq 1): && \int_x^1 \frac{1}{t} \d t &\leq \int_x^1 \frac{1}{t^2} \d t \\ \Rightarrow && \ln 1 - \ln x & \leq -1 + \frac{1}{x} \\ \Rightarrow && 1 - \frac1x &\leq \ln x \\ \end{align*}
3. \(\,\) \begin{align*} (1 < y): && \int_1^y \left (1 - \frac1{x} \right)\d x &\leq \int_1^y \ln x \d x \\ \Rightarrow && \left [x - \ln x \right]_1^y & \leq \left [ x \ln x - x\right]_1^y \\ \Rightarrow && y - \ln y - 1 &\leq y \ln y - y +1 \\ \Rightarrow && 2y-2 & \leq (y+1) \ln y \\ \Rightarrow && \frac{2}{y+1} & \leq \frac{\ln y}{y-1} \\ (0 < y < 1): && \int_y^1 \left (1 - \frac1{x} \right)\d x &\leq \int_y^1 \ln x \d x \\ \Rightarrow && \left [x - \ln x \right]_y^1 & \leq \left [ x \ln x - x\right]_y^1 \\ \Rightarrow && 1 - (y - \ln y) &\leq -1-(y \ln y-y) \\ \Rightarrow && 2-2y &\leq -(y+1)\ln y \\ \Rightarrow && \frac{2}{y+1} &\leq \frac{-\ln y}{1-y} \tag{\(1-y > 0\)} \\ \Rightarrow && \frac{2}{y+1} &\leq \frac{\ln y}{y-1} \\ \\ (1 < y): && \int_1^y \ln x \d x &\leq \int_1^y (x-1) \d x \\ \Rightarrow && \left [x \ln x -x \right]_1^y &\leq \left[ \frac12 x^2 - x \right]_1^y\\ \Rightarrow && y \ln y - y +1 &\leq \frac12y^2 - y+\frac12 \\ \Rightarrow && y \ln y &\leq \frac12 \left (y^2-1 \right) \\ \Rightarrow && \frac{\ln y}{y-1} &\leq \frac{y+1}{2y} \\ \\ (0 < y < 1) && \int_y^1 \ln x \d x &\leq \int_y^1 (x-1) \d x \\ \Rightarrow && \left [x \ln x -x \right]_y^1&\leq \left[ \frac12 x^2 - x \right]_y^1\\ \Rightarrow && -1-(y \ln y - y +1) &\leq-\frac12 - \left ( \frac12y^2 - y\right)\\ \Rightarrow && \frac12 \left (y^2-1 \right) &\leq y \ln y \\ \Rightarrow && \frac{\ln y}{y-1} & \leq \frac{y+1}{2y} \tag{\(y-1 < 0\)} \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. \(\,\) \begin{align*} && I_n &= \int_0^1 x^n \arctan x \d x \\ &&&= \left [ \frac{x^{n+1}}{n+1} \arctan x\right]_0^1 - \int_0^1 \frac{x^{n+1}}{n+1} \frac{1}{1+x^2} \d x \\ &&&= \frac{1}{n+1} \frac{\pi}{4} - \frac{1}{n+1} \int_0^1 \frac{x^{n+1}}{1+x^2}\d x \\ \Rightarrow && (n+1)I_n &= \frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2}\d x \\ && I_0 &= \frac{\pi}{4} - \int_0^1 \frac{x}{1+x^2} \d x \\ &&&= \frac{\pi}{4} - \left [\frac12 \ln(1+x^2) \right]_0^1 \\ &&&= \frac{\pi}{4} - \frac12 \ln 2 \end{align*}
2. \(\,\) \begin{align*} && (n+3)I_{n+2} + (n+1)I_n &=\left ( \frac{\pi}{4} - \int_0^1 \frac{x^{n+3}}{1+x^2} \d x \right)+ \left (\frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2} \d x \right) \\ &&&=\frac{\pi}{2}+ \int_0^1 \frac{x^{n+1}+x^{n+3}}{1+x^2} \d x \\ &&&=\frac{\pi}{2}+ \int_0^1 x^{n+1} \d x \\ &&&= \frac{\pi}{2} + \frac{1}{n+2} \\ && 3I_2 + I_0 &= \frac{\pi}{2} + \frac{1}{2} \\ \Rightarrow && 3I_2 &=\frac{\pi}{4} + \frac12 \ln 2 + \frac12 \\ && 5I_4 + 3I_2 &= \frac{\pi}{2} + \frac14 \\ \Rightarrow && 5I_4 &= \frac{\pi}{2} + \frac14 - \left ( \frac{\pi}{4} + \frac12 \ln 2 + \frac12\right) \\ &&&= \frac{\pi}4-\frac12 \ln 2-\frac14 \\ \Rightarrow && I_4 &= \frac15 \left (\frac{\pi}4-\frac12 \ln 2-\frac14 \right) \\ &&&= \frac1{20} \left (\pi - 2\ln 2 -1 \right) \end{align*}
3. Claim: \[ (4n+1) I_{4n} =\frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2n} (-1)^r \frac 1 {r} \,, \] Proof: Base case we have just shown above Assume true for \(n = k\), consider \(n = k+1\), then \begin{align*} && (4(k+1)+1) I_{4(k+1)} &= \frac{\pi}{2} + \frac{1}{4(k+1)} - (4k+3)I_{4k+2} \\ &&&= \frac{\pi}{2} + \frac{1}{4(k+1)} - \left (\frac{\pi}{2} + \frac{1}{2(2k+1)} - (4k+1)I_{4k} \right)\\ &&&= (4k+1)I_{4k} - \frac12 \left ( \frac{1}{2k+2} - \frac{1}{2k+1}\right) \\ &&&= \frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2k} (-1)^r \frac 1 {r} - \frac12 \left ( \frac{1}{2k+2} - \frac{1}{2k+1}\right)\\ &&&= \frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2(k+1)} (-1)^r \frac 1 {r} \\ \end{align*} as required.

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: The force delivered by the engine must be \(ma + R + Av^2\), (so the net force is \(ma\)). Therefore the work done is \(\displaystyle \int_0^d F \d x = \int_0^d (ma + R + Av^2) \d x\) Notice that \(a = v \frac{\d v}{\d x} \Rightarrow \frac{a}{v} = \frac{\d v}{\d x}\) and so \begin{align*} && WD &= \int_0^d (ma + R + Av^2) \d x \\ &&&= \int_{x=0}^{x=d} (ma + R + Av^2) \frac{v}{a} \frac{\d v}{\d x} \d x \\ &&&= \int_{x=0}^{x=d} \frac{ (ma + R + Av^2)v}{a} \d v \\ \end{align*} Also notice that if we move with constant acceleration from rest for a distance \(d\) the final speed is \(v^2 = 2ad \Rightarrow v = \sqrt{2ad}\)

1. For the second part of the journey, the engine will be putting out a force of \(-ma+R+Av^2>0\), and the car will have a final speed of \(0\) \begin{align*} WD &= \int_0^{w} \frac{(ma+R+Av^2)v}{a} \d v + \int_w^0 \frac{(-ma+R+Av^2)v}{-a} \d v \\ &= \int_0^w \frac{2(Rv+Av^3)}{a} \d v \\ &= \frac{Rw^2+\frac12Aw^4}{a} \\ &= \frac{R2ad+\frac12A4a^2d^2}{a} \\ &= 2Rd + 2Aad^2 \end{align*}
2. If \(ma - 2Aad < R < ma\) then the driving force is still \(-ma+R+Av^2\) which is positive when \(v = \sqrt{2as}\) but negative when \(v = 0\), and therefore at some point in-between the driving force must be \(0\). The engine will stop working when \(-ma+R+Av^2 =0 \Rightarrow v = \sqrt{\frac{ma-R}{A}}\) so \begin{align*} WD &= \int_0^w \frac{(ma+R+Av^2)v}{a} \d v + \int_w^{ \sqrt{\frac{ma-R}{A}}} \frac{(-ma+R+Av^2)v}{-a} \d v \\ &= \int_0^w \frac{2(R+Av^2)v}{a} \d v - \int_0^{\sqrt{\frac{ma-R}{A}}} \frac{(-ma+R+Av^2)v}{a}\d v \\ &= 2Aad^2+2Rd + \frac1a\left (\frac12(R-ma)\frac{ma-R}{A} + \frac{A}{4}\left ( \frac{ma-R}{A}\right)^2 \right) \\ &= 2Aad^2+2Rd - \frac{(ma-R)^2}{Aa}\left (-\frac12+ \frac14 \right) \\ &= 2Aad^2+2Rd - \frac{(ma-R)^2}{4Aa} \end{align*}

Solution:

1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

Solution:

1. \(\,\) \begin{align*} && T(x) &= \int_0^x \! \frac 1 {1+u^2} \, \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u - \int_x^\infty \frac{1}{1+u^2} \d u \\ &&&= T_\infty - \int_x^\infty \frac{1}{1+u^2} \d u \\ u = 1/v, \d u = -1/v^2 \d v: &&&= T_\infty - \int_{v=x^{-1}}^{v=0} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= T_\infty - \int_{0}^{x^{-1}} \frac{1}{1+v^2} \d v \\ &&&= T_\infty - T(x^{-1}) \end{align*}
2. Let \(v = \frac{u+a}{1-au}\) then \begin{align*} && \frac{\d v}{\d u} &= \frac{(1-au) \cdot 1 - (u+a)\cdot(-a)}{(1-au)^2} \\ &&&= \frac{1-au+au+a^2}{(1-au)^2} \\ &&&= \frac{1+a^2}{(1-au)^2} \\ \\ && \frac{1+v^2}{1+u^2} &= \frac{1 + \left ( \frac{u+a}{1-ua} \right)^2}{1+u^2} \\ &&&= \frac{(1-ua)^2+(u+a)^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+u^2a^2+u^2+a^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{(1+u^2)(1+a^2)}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+a^2}{(1-ua)^2} \end{align*} if \(a > 0, x < \frac1a\) then \begin{align*} && T(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_{v=a}^{v=\frac{a+x}{1-ax}} \frac{1}{1+u^2} \frac{1+u^2}{1+v^2} \d v \\ &&&= T\left ( \frac{x+a}{1-ax} \right) - T(a) \\ \\ \Rightarrow && T(x^{-1}) &= T_\infty - T(x) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T(a) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T_\infty-T(a^{-1}) \\ &&&= 2T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) -T(a^{-1}) \end{align*} \(b > 0, y > \frac1b\) then \(y> 0, b > \frac1y\) (same as letting \(x = \frac1y, a = \frac1b\) \begin{align*} && T(y) &= 2T_\infty - 2T \left ( \frac{\frac1y+\frac1b}{1-\frac1{by}} \right) + T(b) \\ \Rightarrow && T(y) &= 2T_\infty - 2T \left ( \frac{b+y}{by-1} \right) + T(b) \\ \end{align*}
3. Letting \(y = b = \sqrt{3}\) in the final equation \begin{align*} && T(\sqrt{3}) &= 2T_{\infty} - T \left ( \frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}\sqrt{3}-1} \right) -T (\sqrt{3}) \\ &&&= 2T_\infty - 2T(\sqrt{3}) \\ \Rightarrow && T(\sqrt{3}) &= \tfrac23 T_\infty \end{align*} Let \(x = \sqrt2 - 1, a = 1\) so, \begin{align*} && T(\sqrt2 -1) &= T \left ( \frac{\sqrt2-1+1}{1-\sqrt2+1} \right)-T(1) \\ &&&= T \left ( \frac{\sqrt{2}}{2-\sqrt{2}} \right) - T(1) \\ &&&= T(\frac{\sqrt{2}(2+\sqrt{2})}{2}) - T(1) \\ &&&= T(\sqrt{2}+1) - T(1) \\ &&&= T_\infty - T(\sqrt2-1)-T(1) \\ \Rightarrow && T(\sqrt{2}-1) &= \frac12T_\infty-\frac12T(1) \\ && T(1) &= T_\infty - T(1) \\ \Rightarrow && T(1) &= \frac12 T_\infty \\ \Rightarrow && T(\sqrt2-1) &= \frac12T_\infty - \frac14T_\infty \\ &&&= \frac14 T_\infty \end{align*}

Solution: \begin{align*} && y &= (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \\ && y' &= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + (ax^2+bx+c) \frac{1}{x + \sqrt{1+x^2}} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}} \right) + d\sqrt{1+x^2} + \frac{x(dx+e)}{\sqrt{1+x^2}} \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (ax^2+bx+c) + d(1+x^2) + x(dx+e) \right) \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (a+2d)x^2+(b+e)x+(d+c) \right) \\ \end{align*}

1. We want \(a = 0, b = 1, d = 0, e = -1, c =0\), so \begin{align*} I &= \int \ln \big( x+\sqrt{1+x^2}\,\big) \,\d x \\ &= x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C \end{align*}
2. We want \(a = b =0, e = 0, d = \frac12, c = \frac12\), so \begin{align*} I &= \int \sqrt{1+x^2}\, \d x \\ &= \frac12\ln(x+\sqrt{1+x^2}) + \frac12x\sqrt{1+x^2}+C \end{align*}
3. We want \(a = \frac12, b = 0, d = -\frac14, e = 0, c = \frac14\), so \begin{align*} I &= \int x \ln (x+\sqrt{1+x^2}) \, \d x \\ &= \left (\frac12 x^2+\frac14 \right)\ln(x+\sqrt{1+x^2}) -\frac14x\sqrt{1+x^2}+C \end{align*}

Solution: \begin{align*} u = a-x, \d u = - \d x: && \int_0^a f(x) \d x &= \int_{u=a}^{u=0} f(a-u) (-1) \d u \\ &&&= \int_0^a f(a-u) \d u \\ &&&= \int_0^a f(a-x) \d x \end{align*}

1. \begin{align*} && I &= \int_0^{\frac12 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\sin (\frac12 \pi - x)}{\cos (\frac12 \pi-x) + \sin (\frac12 \pi-x) } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\cos x}{\sin x + \cos x } \d x\\ \Rightarrow && 2I &= \int_0^{\frac12 \pi} 1 \d x \\ \Rightarrow && I &= \frac{\pi}{4} \end{align*}
2. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\sin (\frac14 \pi - x)}{\cos (\frac14 \pi-x) + \sin (\frac14 \pi-x) } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}{\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x + \frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x} \d x \\ &&&= \int_0^{\frac14 \pi} \frac{\cos x - \sin x}{2 \cos x} \d x \\ &&&= \left [\frac12 x + \ln(\cos x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} -\frac12\ln2 - 1 \end{align*}
3. \begin{align*} && I &= \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \\ &&&= \int_0^{\frac14 \pi} \ln \left (1 + \tan \left(\frac{\pi}{4} - x\right) \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (1 +\frac{1 - \tan x}{1+ \tan x} \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (\frac{2}{1+ \tan x} \right) \, \d x\\ &&&= \frac{\pi}{4} \ln 2 - I \\ \Rightarrow && I &= \frac{\pi}{8} \ln 2 \end{align*}
4. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac x {\cos x \, (\cos x + \sin x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {(\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) \, (\frac{2}{\sqrt{2}}\cos x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {\cos x \, (\cos x + \sin x)}\, \d x \\ \\ \Rightarrow && I &= \frac{\pi}{8} \int_0^{\pi/4} \frac{\sec^2 x}{1 + \tan x} \d x\\ &&&= \frac{\pi}{8} \left [\ln (1 + \tan x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} \ln 2 \end{align*}

Solution: \begin{align*} && \int_{m-\frac12}^\infty \frac{1}{x^2} \d x &= \lim_{K \to \infty} \left [ -x^{-1} \right]_{m-\frac12}^K \\ &&&= \frac{1}{m-\frac12} - \lim_{K \to \infty }\frac{1}K \\ &&&= \frac{1}{m-\frac12} \end{align*}

+ - ↺

1. \(\,\) \begin{align*} E &\approx \int_{1-\frac12}^\infty \frac1{x^2} \d x = \frac{1}{1-\frac12} = 2 \\ E &\approx 1 + \int_{2-\frac12}^\infty \frac1{x^2} \d x= 1 + \frac{1}{2 - \frac12} = \frac53 \\ E &\approx 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x= \frac54 + \frac{1}{3-\frac12} \\ &= \frac54+\frac{2}{5} = \frac{33}{20} \end{align*}
2. The error is \begin{align*} && \epsilon &= \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x - \frac1{r^2} \\ &&&= \frac{1}{r-\frac12} - \frac{1}{r + \frac12} - \frac1{r^2} \\ &&&= \frac{1}{r^2 - \frac14} - \frac1{r^2} \\ &&&= \frac{\frac14}{r^2(r^2-\frac14)} \\ &&&\approx \frac{1}{4r^4} \end{align*} Therefore \begin{align*} && \sum_{n=1}^\infty \frac1{r^4} &\approx 4 \left ( 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x-\sum_{r=1}^\infty \frac{1}{r^2} \right) + 1 + \frac{1}{2^4}\\ &&&= 4 \left ( \frac{33}{20}-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 4 \left ( 1.65-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 1.0825 \approx 1.08 \end{align*}

Solution:

1. \(\,\) \begin{align*} && I_1 &= \int_{-\infty}^{\infty} \frac{1}{x^2+2ax+b} \d x \\ &&&= \int_{-\infty}^{\infty} \frac{1}{b-a^2 +(x+a)^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{b-a^2}} \tan^{-1} \frac{x+a}{\sqrt{b-a^2}} \right]_{-\infty}^{\infty} \\ &&&= \frac{\pi}{\sqrt{b-a^2}} \end{align*}
2. \(\,\) Here is the corrected LaTeX code for the second part, maintaining your exact styling and notation.
3. \(\,\) \begin{align*} && I_{n} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+2ax+b)^{n}} \d x \\ &&&= \left[ \frac{x}{(x^2+2ax+b)^n} \right]_{-\infty}^{\infty} - \int_{-\infty}^\infty x \cdot \frac{-n(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 0 + n \int_{-\infty}^\infty \frac{2x^2+2ax}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= n \int_{-\infty}^\infty \frac{2(x^2+2ax+b) - (2ax+2b)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{2ax+2b}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a) + 2(b-a^2)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x - 2n(b-a^2) I_{n+1} \\ &&&= 2n I_n - n \left[ \frac{-a}{n(x^2+2ax+b)^n} \right]_{-\infty}^\infty - 2n(b-a^2) I{n+1} \\ &&&= 2n I_n - 0 - 2n(b-a^2) I_{n+1} \\ \Rightarrow && 2n(b-a^2)I_{n+1} &= (2n-1)I_n \end{align*}
4. \(\,\) \begin{align*} && I_{n+1} &= \frac{2n-1}{2n(b-a^2)} I_n \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} I_{n-1} \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} \cdots I_{1} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2n \cdot 2(n-1) \cdots 2 (b-a^2)^n} \frac{\pi}{\sqrt{b-a^2}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2^n n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1 \cdot 2n \cdot 2(n-1) \cdots 2}{2^{2n} n!n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n)!}{2^{2n}n!n!}\frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{\pi}{2^{2n}(b-a^2)^{n+\frac12}} \binom{2n}{n} \\ \Rightarrow && I_n &= \frac{\pi}{2^{2n-2}(b-a^2)^{n-\frac12}} \binom{2n-2}{n-1} \\ \end{align*}

Solution:

1. \begin{align*} && \int \frac {x^3-2}{(x+1)^2}\, \e ^x \d x &= \frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{x^3-2}{(x+1)^2}e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \\ \Rightarrow && Q(x)^2(x^3-2) &= ((P(x)+P'(x))Q(x)-Q'(x)P(x))(x+1)^2 \\ \Rightarrow && Q(-1) &= 0 \\ \Rightarrow && x+1 &\mid Q(x) \end{align*} We have \(\frac{x^3-2}{(x+1)^2}\) has degree \(1\) (plus some remainder term). Therefore \begin{align*} 1 &= \deg \l (P(x)+P'(x))Q(x)-Q'(x)P(x)\r - 2 \deg Q(x) \\ &= \deg P(x) + \deg Q(x) - 2 \deg Q(x) \\ &= \deg P(x) - \deg Q(x) \end{align*} as required. Suppose \(Q(x) = x+1, P(x) = ax^2+bx+c\) then \begin{align*} && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))(x+1)-P(x)}{(x+1)^2} \\ \Rightarrow && x^3-2 &= (P(x)+P'(x))(x+1) - P(x) \\ \Rightarrow && x^3-2 &= (ax^2+bx+c+2ax+b)(x+1) - (ax^2+bx+c) \\ &&&= a x^3+ x^2 (2 a + b) + x (2 a + b + c)+b \\ \Rightarrow && a &= 1 \\ && b &= -2 \\ && c &= 0 \end{align*} So \(P(x) = x^2-2x\)
2. \begin{align*} && \int \frac1{x+1}e^x \d x &= \frac{P(x)}{Q(x)}e^x + c \\ \Rightarrow && \frac{1}{x+1} e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{1}{x+1} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \end{align*} Therefore \(Q(-1) = 0\) and so \(x +1 \mid Q(x)\). Considering degrees, we must have that \(P(x)\) has degree \(1\) less than \(Q(x)\). Consider also the number of factors of \(x+1\) in the numerator and denominator. Since \(P(x)\) and \(Q(x)\) have no common factors, the \(Q(x)\) could have \(q\) factors and \(P(x)\) must have none. The denominator therefore has \(2q\) factors and the numerator must have \(q-1\) factors (coming from \(Q'(x)\)), we must have \(2q = (q-1) + 1\), but that implies \(q = 0\). Contradiction! \end{align*}

Solution:

1. \(\,\) \begin{align*} && F_{res} &= kv \\ && P &= Fv \\ v = 4U: && 0 &= F-F_{res} \\ \Rightarrow && 0 &= \frac{P}{4U} - 4Uk \\ \Rightarrow && k &= \frac{P}{16U^2} \\ \\ &&m v \frac{\d v}{\d x}&= \frac{P}{v} - \frac{P}{16U^2}v \\ \Rightarrow && X_1 &= \int_{v=U}^{v=2U} \frac{16U^2mv^2}{P(16U^2-v^2)} \d v \\ v = Ut&& &= \frac{16mU^2}{P} \int_{t=1}^{t=2}\left ( \frac{t^2}{16-t^2} \right)U\d t \\ &&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 + \frac{16}{16-t^2} \right) \d t \\ &&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 +\frac{2}{4+t} +\frac{2}{4-t} \right) \d t \\ &&&= \frac{1}{\lambda}\left (-1 + 2\ln(6)-2\ln(2)-2\ln(5)+2\ln(3) \right) \\ \Rightarrow && \lambda X_1 &= 2\ln \tfrac95-1 \end{align*}
2. \(\,\) \begin{align*} && F_{res} = kv^2 \\ v = 4U: && 0 &= \frac{P}{4U} - 16U^2k \\ \Rightarrow && k &= \frac{P}{64U^3} \\ \\ && mv \frac{\d v}{\d x} &= \frac{P}{v} - \frac{P}{64U^3}v^2 \\ \Rightarrow && X_2 &= \int_{v=U}^{v=2U} \frac{64U^3mv^2}{P(64U^3-v^3)} \d v \\ &&&= \frac{64U^3m}{P} \int_{v=U}^{v=2U} \frac{v^2}{64U^3-v^3} \d v\\ v = Ut &&&= \frac{64U^3m}{P} \int_{t=1}^{t=2} \frac{U^2t^2}{64U^3-U^3v^3} U \d t\\ &&&= \frac{4}{\lambda} \int_1^2 \frac{t^2}{64-t^3} \d t \\ &&&= \frac{4}{\lambda} \left [ -\frac13\ln(64-t^3) \right]_1^2 \\ &&&= \frac{4}{3\lambda} \ln (63/56) \\ \Rightarrow && \lambda X_2 &= \tfrac43 \ln \tfrac98 \end{align*}
3. \(\,\) \begin{align*} && 2\ln \tfrac95 - 1 &\overset{?}{>} \frac43 \ln \frac98 \\ \Leftrightarrow && 4 \ln 3 - 2\ln 5 - 1 &\overset{?}{>} \frac83\ln 3 -4 \ln 2 \\ \Leftrightarrow && \frac43(3\ln 3 + 3\ln 2 - 2 \ln 3) &\overset{?}{>} 2 \ln 5 + 1\\ \Leftrightarrow && \frac43\ln 24 &\overset{?}{>} 2 \ln 5 + 1\\ \end{align*} The \(LHS\) is \(>4.22\). The \(RHS\) is \(< 4.22\), and therefore our inequality holds, in particular, \(X_1 > X_2\).

Solution: \begin{align*} &&\kappa_{X-a} &= \frac{\mathbb{E}\left(\left(X-a-(\mu-a)\right)^4\right)}{\sigma_{X-a}^4}-3 \\ &&&= \frac{\mathbb{E}\left(\left(X-\mu\right)^4\right)}{\sigma_X^4}-3\\ &&&= \kappa_X \end{align*}

1. \(\,\) \begin{align*} && \kappa &= \frac{\mathbb{E}((X-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\mu+\sigma Z-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\sigma Z)^4)}{\sigma^4} - 3 \\ &&&= \mathbb{E}(Z^4)-3\\ &&&= \int_{-\infty}^{\infty} x^4\frac{1}{\sqrt{2\pi}} \exp \left ( - \frac12x^2 \right)\d x -3 \\ &&&= \left [\frac{1}{\sqrt{2\pi}}x^{3} \cdot \left ( -\exp \left ( - \frac12x^2 \right)\right) \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 3x^2 \exp \left ( - \frac12x^2 \right) \d x - 3 \\ &&&= 0 + 3 \textrm{Var}(Z) - 3 =0 \end{align*}
2. \(\,\) \begin{align*} && \mathbb{E}(T^4) &= \mathbb{E} \left [\left ( \sum\limits_{r=1}^n Y_r\right)^4\right] \\ &&&= \mathbb{E} \left [ \sum_{r=1}^n Y_r^4+\sum_{i\neq j} 4Y_iY_j^3+\sum_{i\neq j} 6Y_i^2Y_j^2+\sum_{i\neq j \neq k} 12Y_iY_jY_k^2 +\sum_{i\neq j\neq k \neq l}24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+\sum_{i\neq j} \mathbb{E} \left [ 4Y_iY_j^3\right]+\sum_{i\neq j} \mathbb{E} \left [ 6Y_i^2Y_j^2\right]+\sum_{i\neq j \neq k} \mathbb{E} \left [ 12Y_iY_jY_k^2\right] +\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ 24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+4\sum_{i\neq j} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j^3\right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]+12\sum_{i\neq j \neq k} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k^2\right] +24\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k]\mathbb{E}[Y_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right] \end{align*}
3. Without loss of generality, we may assume they all have mean zero. Therefore we can consider the sitatuion as in the previous case with \(T\) and \(Y_i\)s. Note that \(\mathbb{E}(Y_i^4) = \sigma^4(\kappa + 3)\) and \(\textrm{Var}(T) = n \sigma^2\) \begin{align*} && \kappa_T &= \frac{\mathbb{E}(T^4)}{(\textrm{Var}(T))^2} - 3 \\ &&&= \frac{\sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2\right]\mathbb{E}\left[Y_j^2\right]}{n^2\sigma^4}-3 \\ &&&= \frac{n\sigma^4(\kappa+3)+6\binom{n}{2}\sigma^4}{n^2\sigma^4} -3\\ &&&= \frac{\kappa}{n} + \frac{3n + \frac{6n(n-1)}{2}}{n^2} - 3 \\ &&&= \frac{\kappa}{n} + \frac{3n^2}{n^2}-3 \\ &&&= \frac{\kappa}{n} \end{align*}

Solution: At the point \((2t, t^2)\) the gradient is \(t\). Suppose \(\tan \theta = t\), then the point \(b\) away in each direction is \(\binom{2t}{t^2} \pm b \binom{\cos \theta}{\sin \theta}\), ie one end can be written in the form \((x,y) = (2\tan \theta - b \cos \theta, \tan^2 \theta - b \sin \theta)\). Notice we must have \(2\tan \alpha- b \cos \alpha= 0 \Rightarrow b = 2 \frac{\sin \alpha}{\cos ^2 \alpha}\), therefore the coordinates are \((2 \tan \alpha - 2 \tan \alpha, \tan^2 \alpha - 2\tan^2 \alpha) = (0, -\tan^2 \alpha)\) and \((4 \tan \alpha, 3\tan^2 \alpha)\)

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