Integrals

Multiply both the numerator and denominator by \(\sec^2 x\).

\begin{align} & I =\int_{0}^{\pi / 2} \frac{\mathrm{sec}^{2} x \mathrm{d} x}{(\sqrt{\tan x}+1)^{4}} \\\\ \mathrm{let} \tan x=t^{2} & \Rightarrow \sec^2 x \mathrm{d} x=2 t\mathrm{d}t \\\\ \Rightarrow \quad & I =\int_{0}^{\infty} \frac{2t \mathrm{d}t}{(t+1)^{4}} \Rightarrow I =\int_{0}^{\infty}\left[\frac{2}{(t+1)^{3}}-\frac{2}{(t+1)^{4}}\right] \mathrm{d}t \\ \Rightarrow \quad & I=\left.\frac{-1}{(\mathrm{t}+1)^{2}}+\frac{2}{3(\mathrm{t}+1)^{3}}\right\rvert_{0}^{\infty} \\ \Rightarrow \quad & I=\frac{1}{3} \end{align}

(a) Due to symmetry, we can double the integral from 0 to 3. So the value is \(2I\), where:

\begin{align} I &=\int_{0}^{1} (3-3x^{2}) \text{d}x + \int_{1}^{3} (3 x^{2}-3) \text{d}x \\ &=\left. 3x-x^3 \right\rvert_{0}^1 + \left. (x^{3}-3x)\right\rvert_{1}^3 \\ &=2 + 20 \\ &=22\\ \end{align} Hence, \(\int_{-3}^{3}\left|3 x^{2}-3\right| d x = 44 \).

(b) \(\left|3 x^{2}-3\right|\) is a continuous function so by the fundamental theorem of calculus, \(f^{\prime}(1)=\) \(\left|3 \times 1^{2}-3\right|=0\)

(a) \begin{align} \int_1^{e^2} \ln x \: \mbox{d}x &= x\ln x - x \rvert_{1}^{e^2} \\ &= [e^2\ln e^2 - e^2] -[ 1\ln 1 - 1] \\ &= e^2+1 \end{align}

(b) Since \(|x| = |-x|\), we have: \begin{align} \frac{I}{2} &= \int_{0}^1 \frac{\ln x}{x} \mbox{d}x \\ &= \left. \frac{(\ln x)^2}{2} \right \rvert_{0}^{1} \\ &= \frac{(\ln 1)^2}{2} - \frac{ (\ln 0)^2 }{2} \\ &=0-\infty \\ \therefore I &= -\infty \end{align}

(a) The volume of the cave is given by the following triple integral:

\[ \int_{x=0}^{1} \int_{y=-\sqrt{1-x}}^{\sqrt{1-x}} \int_{z=0}^{2\sqrt{1-x}} \mathrm{d}x \mathrm{d}y \mathrm{d}z \] (b) \[ \int_{x=0}^{1} \int_{y=-\sqrt{1-x}}^{\sqrt{1-x}} 2\sqrt{1-x}\; \mathrm{d}x \mathrm{d}y \] \[ \int_{x=0}^{1} 4(1-x) \mathrm{d}x = 2 \text{m}^3 \]

Here is an alternate way to find the volume: The area of the cross-section of the cave goes from 4 sq.m to 0 sq.m linearly as \(x\) goes from 0 to 1. So volume is \(\frac{1}{2} \times 1 \times 4 = 2 \text{m}^3\).

Put \(x=\tan u\). This changes the integral to \[ I= \int_{0}^{\pi/2}\cos^{2m}(u) du \] \[ I=\frac{1}{4} \int_{0}^{2 \pi}\cos^{2m}(u) du \] \[\begin{aligned} \cos ^{n} \theta &=2^{-n}\left(\mathrm{e}^{\mathrm{i} \theta}+\mathrm{e}^{-\mathrm{i} \theta}\right)^{n} \\ &=2^{-n} \sum_{k=0}^{n}\left(\begin{array}{l}n \\ k\end{array}\right) \mathrm{e}^{\mathrm{i} k \theta} \mathrm{e}^{-\mathrm{i}(n-k) \theta} \\ &=2^{-n} \sum_{k=0}^{n}\left(\begin{array}{l}n \\ k\end{array}\right) \mathrm{e}^{\mathrm{i}(2 k-n) \theta} \\ \cos^{2m}\theta &=2^{-2m} \sum_{k=0}^{2m} \binom{2m}{k} \mathrm{e}^{\mathrm{i}(2k-2m) \theta \quad\quad\quad (1) } \end{aligned}\] \[\text{ Since } \int_{0}^{2 \pi} e^{i \ell x} d x=\left\{\begin{array}{ll}2 \pi & (\ell=0) \\ 0 & (\ell \neq 0)\end{array}\right.\] only one term in \((1)\) is non-zero, when \(k=m\). \[\int_{0}^{2 \pi} \cos ^{2m} x d x=\frac{2 \pi}{2^{2m}} \binom{2m}{m} \] \[I = \frac{1}{4} \cdot \frac{2\pi}{2^{2m}} \cdot \binom{2m}{m}\]

There is a mistake in this problem. The above statement is not true. See this discussion on Math Stack Exchange for details.

a) Let \(f(t)=1 /\left(1+t^{2}\right) .\) Then \(y\) and \(x\) are respectively the left and right hand Riemann sums for \(f\) over the interval \(\left[0, \frac{1}{\sqrt{3}}\right]\) using 10 equal parts, each of width \(1 / 10 \sqrt{3} .\) since \(f\) is a positive decreasing function, \(y\) must be greater than the area under \(f\) over the given interval and \(x\) must be lesser than the area. The area under \(f\) over \(\left[0, \frac{1}{\sqrt{3}}\right]\) is:

\[\int_{0}^{1 / \sqrt{3}} f(t) d t=\lvert \tan^{-1}(t)\rvert_{0}^{1 / \sqrt{3}}=\frac{\pi}{6}\]

This implies that \(x< \frac{\pi}{6}< y\).

b) \(\frac{x+y}{2}\) can be interpreted as the sum of areas of 10 trapezoids as follows. Dividing \(\left[0, \frac{1}{\sqrt{3}}\right]\) into 10 equal parts, let the \(i\)th subinterval be \(\left[x_{i}, x_{i+1}\right]\) with \(i=0,1, \ldots, 10 .\) Then the \(i\) -th trapezoid has base \(\left[x_{i}, x_{i+1}\right]\) and it has two vertical sides, the left one of height \(f\left(x_{i}\right)\) and the right one of height \(f\left(x_{i+1}\right)\).

Claim: The function \(f\) is concave

Proof: In the interval \(\left(0, \frac{1}{\sqrt{3}}\right)\), we have \(f^{\prime \prime}(t)=\frac{6 t^{2}-2}{\left(1+t^{2}\right)^{3}}<0\). \(\square\).

Since \(f\) is concave, the total area of trapezoids is less than the area under \(f\).

a) \(\int_{1}^{4} x^{2} d x=\frac{x^{3}}{3}\rvert_{1} ^{4}=21\) using the fundamental theorem of calculus.

b) \(\int_{1}^{3}[x]^{2} d x=1\left(1^{2}\right)+1\left(2^{2}\right)=5=\) area under the piecewise constant function \([x]^{2}\)

c) \(\int_{1}^{2}\left[x^{2}\right] d x=1(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})=5-\sqrt{2}-\sqrt{3}\) since the function \([x]^{2}\) is constant on intervals \([1, \sqrt{2}),[\sqrt{2}, \sqrt{3}),[\sqrt{3}, 2)\). It takes the values 1,2 and 3 on these intervals, respectively.

d) \(\int_{-1}^{1} \frac{1}{x^{2}} d x=2 \lim_{t \rightarrow 0^{+}} \int_{t}^{1} \frac{1}{x^{2}} d x=2 \lim_{t \rightarrow 0^{+}}\left(-1+\frac{1}{t}\right)=\infty .\) The fundamental theorem does not apply over the interval [-1,1] because \(\frac{1}{x^{2}}\) goes to \(\infty\) in the interval. So the integral does not exist.

(i) Yes. We must define \(g(0)=\lim_{x \rightarrow 0} g(x)=f^{\prime}(0),\) which exists by hypothesis.

(ii) Consider \(\int_{-1}^{-r} g(x) d x=\int_{-1}^{-r} \frac{f(x)}{x} d x-\int_{-1}^{-r} \frac{f(0)}{x} d x=\int_{-1}^{-r} \frac{f(x)}{x} d x-f(0) \ln r . \quad\) Similarly

\[\int_{r}^{1} g(x) d x=\int_{r}^{1} \frac{f(x)}{x} d x-\int_{r}^{1} \frac{f(0)}{x} d x=\int_{r}^{1} \frac{f(x)}{x} d x+f(0) \ln r\]

Thus the expression inside the given limit is equal to \(\int_{-1}^{-r} g(x) d x+\int_{r}^{1} g(x) d x,\) as \(\pm f(0) \ln r\) cancels out.

Applying the fundamental theorem of calculus to the continuous function \(g,\) we get an anti-derivative \(G\) of \(g,\) where \(G\) is defined on [-1,1] by \(G(t)=\int_{-1}^{t} g(x) d x .\) So the given

\begin{align} \text { limit }&=\lim_{r \rightarrow 0^{+}}\left(\int_{-1}^{-r} g(x) d x+\int_{r}^{1} g(x) d x\right) \\ &=\lim_{r \rightarrow 0^{+}}(G(-r)-G(-1)+G(1)-G(r)) \\ &=G(0)-0+G(1)-G(0)\\ &=G(1) \end{align}
We have used the fundamental theorem to evaluate the integrals and the fact that \(G\) is differentiable.

(iii) Define \(f(x)=\frac{1}{-\ln \frac{x}{2}}\) for \(x \in(0,1], f(x)=\frac{1}{\ln \left|\frac{1}{2}\right|}\) for \(x \in[-1,0),\) and \(f(0)=0 .\) Verify that this works: \(f\) is continuous at 0 and so on \([-1,1] .\) It is not differentiable at 0 as the relevant limit is \(+\infty .\) The two integrals in the desired limit are equal (because \(f\) is an odd function, so \(\frac{f(x)}{x}\) is even \()\) and each integral is \(+\infty\) as it amounts to \(\lim_{t \rightarrow 0^{+}} \ln |\ln t| .\) Can you see how one might think of such \(f ?\) E.g., check that choices like \(|x|\) or even \(x^{\frac{1}{3}}\) do not work. Compare the behavior of these functions at \(x=0\) with that of chosen \(f .\)

(a) False. \( f(x) = 0 \) is a counterexample.

(b) False. There are infinitely many candidates for \(f\). The maximum value of \( f(x)(1-f(x)) \) in the interval \(x \in (0,1)\) is \(1/4\). This value is attained only when \(f(x)=1/2\). So \(f\) is unique in the interval \( (0,1) \). But \(f\) is defined on \(\mathbb{R}\), so outside the interval \((0,1)\), it can be any continuous function.

(c) True. See part (b).

(a) Note that the given inequality stays valid if we take absolute value of the LHS, because we may interchange \(x\) and \(y\) without affecting RHS.

Fix \(x, y\) and let \(t=x-y .\) For now let \(x \geq y,\) i.e. \(t \geq 0 .\) For \(h \in[0, t],\) the value of \(y+h\) varies between \(y\) and \(x .\) We are given that \(\left|f^{\prime}(y+h)-f^{\prime}(y)\right| \leq 3|h|\). Integrate with respect to \(h\) over the interval \([0, t]\) to get: \[\int_{0}^{t}\left|f^{\prime}(y+h)-f^{\prime}(y)\right| d h \leq \int_{0}^{t} 3|h| d h=1.5 t^{2}\]
The following inequality follows from Lemma 1 (proved below).

\[\left|\int_{0}^{t} f^{\prime}(y+h)-f^{\prime}(y) d h\right| \leq \int_{0}^{t}\left|f^{\prime}(y+h)-f^{\prime}(y)\right| d h .\]

Combining with the previous inequality we have: \[\left|\int_{0}^{t} f^{\prime}(y+h)-f^{\prime}(y) d h\right| \leq 1.5 t^{2}\]

Finally we calculate the LHS.
\[\left|\int_{0}^{t} f^{\prime}(y+h) d h-\int_{0}^{t} f^{\prime}(y) d h\right|=\left|f(y+t)-f(y)-f^{\prime}(y) t\right|\]
where the first integral is calculated using the fundamental theorem of calculus and the second one is just the integral of the constant \(f^{\prime}(y) .\) Substituting \(x-y\) for \(t\) gives the desired result.

The case when \(x\leq y\) is analogous.

(b) We have, for \(x \neq y,\left|\frac{f^{\prime}(x)-f^{\prime}(y)}{x-y}\right| \leq 3,\) so \(-3 \leq \frac{f^{\prime}(x)-f^{\prime}(y)}{x-y} \leq 3 .\) Taking limit as \(y \rightarrow x\) we get \(-3 \leq f^{\prime \prime}(x) \leq 3 .\) It is easy to provide examples where \(f^{\prime \prime}\) attains the extreme values \(\pm 3,\) e.g. \(f(x)=\pm 1.5 x^{2} .\) These satisfy the hypothesis and have constant \(f^{\prime \prime}=\pm 3\)

Lemma 1.

\[ \left|\int_{a}^{b} f(x) d x\right| \leq \int_{a}^{b}|f(x)| d x \]

Proof.

\[ -|f(x)| \leq f(x) \leq|f(x)| \]

for all \(x\); hence

\[ -\int_{a}^{b}|f(x)| d x \leq \int_{a}^{b} f(x) d x \leq \int_{a}^{b}|f(x)| d x \]

The lemma follows since \(-a \leq x \leq a \implies |x|\leq a \) if \(a \geq 0\). \(\quad \square\)

(a) The general form of Leibniz rule is:
\[\frac{\mathrm{d}}{\mathrm{d} x} \int_{a(x)}^{b(x)} f(x, t) \mathrm{d} t=f(x, b(x)) \cdot b^{\prime}(x)-f(x, a(x)) \cdot a^{\prime}(x)+\int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, t) \mathrm{d} t\]

Applying the rule above gives the following answer: \(e^{x} x \cos^{4}(e^{x})\)

(b)

\begin{aligned} F^{\prime}(x) &=\frac{d}{d x} \int_{1}^{x} t \log t \mathrm{d} t \\ &=x \log x \end{aligned}

Therefore \(F^{\prime}(1)=0 .\) Moreover, \(F^{\prime \prime}(x)=1+\log x .\) Hence one concludes that \(F\) is decreasing on \((0,1),\) increasing on \((1, \infty)\) and has a local minima at \(x=1\)

From \(x=0\) to \(x=1\) we have \(\sqrt{x} \geq|-x|,\) so from \(x=0\) to \(x=1\) the volume swept out by the curve determines the volume. From \(x=1\) to \(x=4\), the volume swept by the line determines the volume. Hence, we need two integrals.

From \(x=0\) and \(x=1\) we calculate volume obtained by revolving the area below \(y=\sqrt{x}\). \[ \int_{0}^{1} \pi (\sqrt{x})^2 \text{d}x = \left.\frac{\pi x^2}{2} \right\rvert_{0}^{1} = \frac{\pi\cdot 1^2}{2} - \frac{\pi\cdot 0^2}{2} = \frac{\pi}{2} \] Similarly, from \(x=1\) to \(x=4\) we have \(|-x| \geq \sqrt{x},\) so here we just have to take the volume obtained by the revolving the area below \(y=x\). \[ \int_{1}^{4} \pi x^2 \text{d}x = \left.\frac{\pi x^3}{3} \right\rvert_{1}^{4} = \frac{64\pi}{3} - \frac{\pi}{3} = 21\pi \] Thus total volume of the resulting solid: \( 21.5 \pi \).

1. We split the function into three parts: \(A\) is the interval before the first time the function touches zero. \(B\) is the part of the function that is positive for \(x>\pi/2\) and \(C\) is the part of the function that is negative. We show that each of these intergrals converge.

   

   \begin{align*} A:&\int_{1/x}^{\pi/2} \frac{\cos t}{t^2} dt\\ B:& \sum_{k=0}^{n} \int_{(2k+1)\pi/2}^{(2k+3)\pi/2} \frac{\cos t}{t^2} dt\\ C:& \sum_{k=0}^{n} \int_{(2k+3)\pi/2}^{(2k+5)\pi/2} \frac{\cos t}{t^2} dt\\ \end{align*} \(A\) is a proper integral and hence well-defined for a given \(x\). \(B\) and \(C\) are dominated by the integrals \( \int_{\pi/2}^{\infty} \frac{1}{t^2}\) and \( -\int_{\pi/2}^{\infty} \frac{1}{t^2} dt \), respectively and hence they converge.
2. The general form of Leibniz rule is:
   \[\frac{\mathrm{d}}{\mathrm{d} x} \int_{a(x)}^{b(x)} f(x, t) \mathrm{d} t=f(x, b(x)) \cdot b^{\prime}(x)-f(x, a(x)) \cdot a^{\prime}(x)+\int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, t) \mathrm{d} t\] Substituting, \( a(x) = 1/x \) and \(b(x)=L\), we get:
   \begin{align*} f^\prime(x) &= 0- \frac{\cos(1/x)}{1/x^2}\cdot (-1/x^2) \\ f^\prime(1/\pi) &= \cos\pi = -1 \\ \end{align*}
3. Solution due to Soham Ghosh (West Bengal).
   \begin{align*} f(x) &=\int_{1 / x}^{\infty} \frac{1}{t^{2}} \cos t \,dt \\ &=\left.\frac{1}{t^{2}} \sin t\right|_{1 / x} ^{\infty}-\int_{1 / x}^{\infty} \frac{(-2)}{t^{3}} \sin t\, dt \;\;\text{ (using integration by parts)} \\ &=-x^{2} \sin \frac{1}{x}+\int_{1 / x}^{\infty} \frac{2}{t^{3}} \sin t\, dt \\ \end{align*} We are required to calculate only the right-hand side limit. L'Hospital's rule is not applicable. \[ \lim_{h \rightarrow 0^{+}} \frac{f(h)-f(0)}{h}=\lim_{h \rightarrow 0^{+}} \frac{f(h)}{h} \] \begin{align*} f(h)&=-h^{2} \sin \frac{1}{h}+\int_{1 / h}^{\infty} \frac{2}{t^{3}} \sin t\, d t \\ \frac{f(h)}{h}&=-h \sin \frac{1}{h}+\frac{1}{h}\int_{1 / h}^{\infty} \frac{2}{t^{3}} \sin t\, d t \end{align*} We will show that the limit of each summand in the LHS goes to zero.
   Firstly, \( \lim_{h\rightarrow 0} -h \sin \frac{1}{h} = 0 \).
   \begin{align*} -\frac{2}{t^{3}} & \leq \frac{2}{t^{3}} \sin t \leq \frac{2}{t^{3}} \quad (\because-1 \leqslant \sin t \leqslant 1) \\ -\int_{1 / x}^{\infty} \frac{2}{t^{3}} d t &\leq \int_{1 / x}^{\infty} \frac{2}{t^{3}} \sin t\, dt \leq \int_{1 / x}^{\infty} \frac{2}{t^{3}} d t \\ -\left.2 \frac{t^{-2}}{-2}\right|_{1 / x} ^{\infty} &\leq \int_{1 / x}^{\infty} \frac{2}{t^{3}} \sin t\, dt \leq \left. 2 \frac{t^{-2}}{-2}\right|_{1 / x} ^{\infty} \\ -x^{2} &\leq \int_{1 / x}^{\infty} \frac{2}{t^{3}} \sin t\, d t \leq x^{2} \\ -h^{2} &\leq \int_{1 / h}^{\infty} \frac{2}{t^{3}} \sin t\, d t \leq h^{2} \\ -h &\leq \frac{1}{h} \int_{1 / h}^{\infty} \frac{2}{t^{3}} \sin t\, d t \leq h \\ \lim_{h\rightarrow 0^+}& \frac{1}{h} \int_{1 / h}^{\infty} \frac{2}{t^{3}} \sin t\, d t = 0 \;\; \text{ Hence proved.} \end{align*}

(a) It is easy to see that for \(0 \leq x \leq 1,\) we have \(1 \leq \cos (x)+\sin (x) \leq \sqrt{2},\) and so

\[ x^{1} \geq x^{\cos (x)+\sin (x)} \geq x^{\sqrt{2}} \]

As all three functions are non-negative in \([0,1],\) we can integrate the inequalities over that interval to get

\[ \frac{1}{2} \geq \int_{0}^{1} f(x) d x \geq \frac{1}{\sqrt{2}+1}>\frac{1}{1.5+1}=0.4 \]

(b) Length of the curve from \(x=0\) to any given \(x\) is \(l(x)=\int_{0}^{x} \sqrt{1+f^{\prime}(u)^{2}} d u\) It is given that \(\frac{d l}{d t}=1 \mathrm{cm} /\) second at all times. One needs to find \(\frac{d x}{d t}\) when \(x=1\)

By chain rule \( \displaystyle \frac{d l}{d t}=\frac{d l}{d x} \frac{d x}{d t} .\)

By the fundamental theorem of calculus:

\[\frac{d l}{d x}=\sqrt{1+f^{\prime}(x)^{2}}\]

From Lemma 1, we have \(f^{\prime}(1)=\cos (1)+\sin (1)\).

So at \(x=1\) we have

\[ \frac{d l}{d x}=\sqrt{1+(\cos (1)+\sin (1))^{2}}=\sqrt{2+\sin 2}\] The statement follows by substituting in the first equation.

Lemma 1: \(f^{\prime}(1) = \sin(1) + \cos(1)\)

\[f^{\prime}(x) = x^{(\cos (x)+\sin (x))-1}(-x \log (x) \sin (x)+x \log (x) \cos (x)+(\cos (x)+\sin (x))) \]

Substitute \(x=1\) and \(\log (1) = 0\) in the above equation.\(\;\square\)