HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

juantheron · Oct 24, 2015

Re: MX2 2015 Integration Marathon

$(a)\;\;Evaluation of $\displaystyle \bf{\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)}+e^{\tan^{-1}(\cos x)}}dx}$

$(b)\;Determine positive integer $n\leq 5,$ such that $\int_{0}^{1}e^x\cdot (x-1)^ndx = 16-6e$

omegadot · Oct 25, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$(b)\;Determine positive integer $n\leq 5,$ such that $\int_{0}^{1}e^x\cdot (x-1)^ndx = 16-6e$

$Let $ I_n = \int^1_0 {\rm{e}}^x ( x - 1)^n \, dx, \quad n \geqslant 0.\\$On integrating by parts, we have$

$\begin{align*}I_n &= \left [{\rm{e}}^x (x - 1)^n \right ]^1_0 - \int^1_0 {\rm {e}}^x n (x - 1)^{n - 1} \, dx\\&= -(-1)^n - n \int^1_0 {\rm {e}}^x (x - 1)^{n - 1} ]\, dx\\&= (-1)^{n + 1} - n I_{n - 1}, \quad n \geqslant 1.\end{align*}$

$$Finding $I_0.\\\begin{align*}I_0 = \int^1_0 {\rm{e}}^x \, dx = e - 1.\end{align*}$

$$From the recurrence relation, we have $\\\begin{align*}I_1 &= (-1)^2 - I_0 = 1 - (e - 1) = 2 - e\\I_2 &= (-1)^3 - 2 I_1 = -1 - 2 (2 - e) = 2e - 5\\I_3 &= (-1)^4 - 3 I_2 = 1 - 3(2e - 5) = 16 - 6e\end{align*}\\$So the required positive integer $n \leqslant 5$ we seek is $n = 3.$

omegadot · Oct 25, 2015

Re: MX2 2015 Integration Marathon

$\int^\pi_0 \frac{d\theta}{(2 - \cos \theta)^2}$

omegadot · Oct 25, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$(a)\;\;Evaluation of $\displaystyle \bf{\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)}+e^{\tan^{-1}(\cos x)}}dx}$

$$Let $I = \int^{\frac{5\pi}{2}}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\sin x)}}{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}} \, dx.$

$$Now $\\\begin{align*}I &= \int^{0}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\sin x)}}{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}} \, dx + \int^{\frac{5\pi}{2}}_{0} \frac{{\rm{e}}^{\tan^{-1}(\sin x)}}{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}} \, dx.\end{align*}$

$$We intend to make use of the following result for definite integrals $\\\int^b_a f(x) \, dx = \int^b_a f(b + a - x) \, dx.$

$$If $f(x) = \frac{{\rm{e}}^{\tan^{-1}(\sin x)}}{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}},$ then$\\\begin{align*}f(b + a - x) &= f(\pi/2 - x)\\&= \frac{{\rm{e}}^{\tan^{-1}(\sin (\pi/2-x))}}{{\rm{e}}^{\tan^{-1}(\sin (\pi/2 - x))} + {\rm{e}}^{\tan^{-1}(\cos (\pi/2-x))}}\\&= \frac{{\rm{e}}^{\tan^{-1}(\cos x)}}{{\rm{e}}^{\tan^{-1}(\cos x)} + {\rm{e}}^{\tan^{-1}(\sin x)}}\end{align*}\\$were we have made use of the following results\\\begin{align*}\sin \left (\frac{\pi}{2} - x \right ) = \cos x \quad \mbox{and} \quad \cos \left (\frac{\pi}{2} - x \right ) = \sin x.\end{align*}$

$$Similarly$\\\begin{align*}f(b + a - x) &= f(5\pi/2 - x)\\&= \frac{{\rm{e}}^{\tan^{-1}(\sin (5\pi/2-x))}}{{\rm{e}}^{\tan^{-1}(\sin (5\pi/2 - x))} + {\rm{e}}^{\tan^{-1}(\cos (5\pi/2-x))}}\\&= \frac{{\rm{e}}^{\tan^{-1}(\cos x)}}{{\rm{e}}^{\tan^{-1}(\cos x)} + {\rm{e}}^{\tan^{-1}(\sin x)}}\end{align*}\\$were we have made use of the following results\\\begin{align*}\sin \left (\frac{5\pi}{2} - x \right ) = \cos x \quad \mbox{and} \quad \cos \left (\frac{5\pi}{2} - x \right ) = \sin x.\end{align*}$

$$Thus $\\\begin{align*}I &= \int^{0}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\cos x)}}{{\rm{e}}^{\tan^{-1}(\cos x)} + {\rm{e}}^{\tan^{-1}(\sin x)}} \, dx + \int^{\frac{5\pi}{2}}_{0} \frac{{\rm{e}}^{\tan^{-1}(\cos x)}}{{\rm{e}}^{\tan^{-1}(\cos x)} + {\rm{e}}^{\tan^{-1}(\sin x)}} \, dx\\&= \int^{\frac{5\pi}{2}}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\cos x)}}{{\rm{e}}^{\tan^{-1}(\cos x)} + {\rm{e}}^{\tan^{-1}(\sin x)}} \, dx\end{align*}$

$$Adding the two results for $I$ together gives$\\\begin{align*}I + I &= \int^{\frac{5\pi}{2}}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\sin x)}}{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}} \, dx + \int^{\frac{5\pi}{2}}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\cos x)}}{{\rm{e}}^{\tan^{-1}(\cos x)} + {\rm{e}}^{\tan^{-1}(\sin x)}} \, dx\\2I &= \int^{\frac{5\pi}{2}}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}}{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}} \, dx\\\Rightarrow I &= \frac{1}{2}\int^{\frac{5\pi}{2}}_{\frac{\pi}{2}} dx\\&= \left [\frac{x}{2} \right ]^{5\pi/2}_{\pi/2}\\&=\pi \end{align*}$

juantheron · Nov 1, 2015

Re: MX2 2015 Integration Marathon

$Let $I(a) = \int_{0}^{\pi}\frac{1}{a-\cos \theta } d\theta\;,$ Now Put $\cos \theta = \frac{1-\tan^2 \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}}\;,$ we get$

$So $I(a) = \int_{0}^{\pi}\frac{\sec^2 \frac{\theta}{2}}{(a-1)+(a+1)\tan^2 \frac{\theta}{2}}d\theta\;,$ Now Put $\tan \frac{\theta}{2} = t\;,$

$Then $\sec^2 \frac{\theta}{2}d\theta = 2dt$ and changing limits, We get$

$Now evaluate above Integral We get$

$I(a) = \frac{2}{(a+1)}\int_{0}^{\infty}\frac{1}{(\frac{a-1}{a+1})+t^2}dt = \frac{\pi}{\sqrt{a^2-1}} = \frac{\pi}{(a^2-1)^{\frac{1}{2}}}$

$Now Differentiate both side w. r to $a\;,$ We get$

$So $I'(a) = \frac{d}{da}\left[\int_{0}^{\pi}\frac{1}{(a-\cos \theta )}d\theta \right] = \frac{d}{da}\left[\frac{\pi}{(a^2-1)^{\frac{1}{2}}}\right]$

$So $I'(a) = -\int_{0}^{\infty}\frac{1}{(a-\cos \theta)^2}d\theta = -\frac{\pi}{2}(a^2-1)^{-\frac{3}{2}}\cdot 2a = -\frac{\pi\cdot a}{(a^2-1)^{\frac{3}{2}}}$

$So we get $\int_{0}^{\infty}\frac{1}{(a-\cos \theta)^2} = \frac{\pi \cdot a}{(a^2-1)^{\frac{3}{2}}}\;,$ Now Put $a=2$

$So we get $\int_{0}^{\infty}\frac{1}{(2-\cos \theta)^2}d\theta = \frac{2\pi}{3\sqrt{3}}$

juantheron · Nov 1, 2015

Re: MX2 2015 Integration Marathon

$(1)\;\; Evaluation of $\int_{0}^{\pi}\frac{x}{\sqrt{1+\sin^3 x}}((3\pi \cos x+4\sin x)\sin^2 x+4)dx =$

$(2)\;\; If $L = \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n}_{k=1}k\ln\left[\frac{n^2+(k-1)^2}{n^2+k^2}\right]$ exists, Then $L=$

$(3)\;\; If $F(n) = \int\frac{x^n}{e^x}dx\;,$ Then value of Definite Integral $\int_{2}^{5}\left[e^{\int\frac{F(1)}{F(2)}dx}\right]^2dx=$

InteGrand · Nov 1, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$(1)\;\; Evaluation of $\int_{0}^{\pi}\frac{x}{\sqrt{1+\sin^3 x}}((3\pi \cos x+4\sin x)\sin^2 x+4)dx =$

$(2)\;\; If $L = \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n}_{k=1}k\ln\left[\frac{n^2+(k-1)^2}{n^2+k^2}\right]$ exists, Then $L=$

$(3)\;\; If $F(n) = \int\frac{x^n}{e^x}dx\;,$ Then value of Definite Integral $\int_{2}^{5}\left[e^{\int\frac{F(1)}{F(2)}dx}\right]^2dx=$

For Question (3), should there be limits on the integral for F(n)?

Paradoxica · Nov 2, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
For Question (3), should there be limits on the integral for F(n)?

From my best guess, no. The intended outside integral is the exponential of a function in x, and F(1), F(2) are functions of x. If there were borders on the index integral, then the problem would become trivial. This is an MX2 Integration Marathon, isn't it?

InteGrand · Nov 2, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
From my best guess, no. The intended outside integral is the exponential of a function in x, and F(1), F(2) are functions of x. If there were borders on the index integral, then the problem would become trivial. This is an MX2 Integration Marathon, isn't it?

$OK well then if it was intended that $F(n)$ was a function of $x$, it would have been better to write it as $F_n (x)$. As it was written as $F(n)$, it looks like a function only dependent on $n$. Also, note that $F_n (x)$ is defined only up to a constant (as is the integral in the exponent of the integral it asks the value of), so the answer should not be a numerical one, but will have arbitrary constants in it.$

InteGrand · Nov 2, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$(1)\;\; Evaluation of $\int_{0}^{\pi}\frac{x}{\sqrt{1+\sin^3 x}}((3\pi \cos x+4\sin x)\sin^2 x+4)dx =$

$(2)\;\; If $L = \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n}_{k=1}k\ln\left[\frac{n^2+(k-1)^2}{n^2+k^2}\right]$ exists, Then $L=$

$(3)\;\; If $F(n) = \int\frac{x^n}{e^x}dx\;,$ Then value of Definite Integral $\int_{2}^{5}\left[e^{\int\frac{F(1)}{F(2)}dx}\right]^2dx=$

$\underline{Question (2)}$

$Given: $L=\lim_{n\to \infty} \frac{1}{n}\sum _{k=1}^n k \ln \left(\frac{n^2 + (k-1)^2}{n^2 + k^2}\right)$.$

$Let $S_n = \sum _{k=1}^n k \ln \left(\frac{n^2 + (k-1)^2}{n^2 + k^2} \right)$.$

$Note that in general,$

$\begin{align*} \sum_{k=1}^n ka_k &= \sum_{k=1}^n a_k + \sum_{k=2}^n a_k + \sum_{k=3}^n a_k + \cdots + \sum_{k=n-1}^n a_k + \sum_{k=n}^n a_k \\ &= \sum_{j=1}^{n} \sum _{k=j} ^n a_k . \end{align*}$

$For our purposes, we have $a_k = \ln \left(\frac{n^2 + (k-1)^2}{n^2 + k^2} \right)$, $S_n = \sum _{k=1}^n k a_k = \sum_{j=1}^{n} \sum _{k=j} ^n a_k$, and $L=\lim_{n\to \infty} \left(S_n \cdot \frac{1}{n}\right)$.$

$Now, we have$

$\begin{align*}\sum _{k=j}^n a_k &= \sum _{k=j}^n \ln \left(\frac{n^2 + (k-1)^2}{n^2 + k^2} \right) \\ &= \ln \left( \prod _{k=j} ^{n} \frac{n^2 + (k-1)^2}{n^2 + k^2}\right) \quad (\text{log laws}) \\ &= \ln \left( \frac{n^2 + (j-1)^2}{n^2 + j^2}\times \frac{n^2 + {j}^2}{n^2 + {(j+1)}^2}\times \frac{n^2 + {(j+1)}^2}{n^2 + {(j+2)}^2} \times \cdots \times \frac{n^2 + (n-2)^2}{n^2 + (n-1)^2}\times \frac{n^2 + (n-1)^2}{n^2 + n^2}\right) \\ &= \ln \left(\frac{n^2 + (j-1)^2}{2n^2} \right) \quad (\text{telescoping product}). \end{align*}$

$So we have $S_n = \sum _{j=1}^n \ln \left(\frac{n^2 + (j-1)^2}{2n^2} \right)$. Therefore, we have $L = \lim _{n \to \infty} \left[ \sum _{j=1}^n \ln \left(\frac{n^2 + (j-1)^2}{2n^2} \right) \cdot \frac{1}{n}\right]$.$

$Now consider the function $f(x) = \ln \left(\frac{1+x^2}{2}\right)$ for $x\in [0,1]$, and consider the integral $I= \int _0 ^1 f(x)\text{ d}x$. Consider the limit of the Left Riemann Sum for this; the interval $[0,1]$ is uniformly partitioned with the points $\left \{ 0,\frac{1}{n},\frac{2}{n},\ldots , \frac{n}{n}=1 \right\}$, and so the limit of the Left Riemann Sum is $\mathcal{R}_\text{left}= \lim _{n\to \infty}\sum _{i=1}^n f(x_i ^*)\Delta x$, where $x_i ^* = \frac{i-1}{n}$ and $\Delta x = \frac{1}{n}$ for all $i=1,2,\ldots, n$. That is,$

$\begin{align*} \mathcal{R}_\text{left} &= \lim_{n\to \infty} \left[\sum _{i=1}^n \ln \left(\frac{1+\left(\frac{i-1}{n} \right)^2}{2}\right)\cdot \frac{1}{n}\right] \\&= \lim_{n\to \infty} \left[\sum _{i=1}^n \ln \left(\frac{n^2+(i-1)^2}{2n^2}\right)\cdot \frac{1}{n}\right]\\ &= L. \end{align*}$

$So $L$ is just the limit of the Left Riemann Sum for the integral $I$. Since $f$ is Riemann integrable, we have that$

$\begin{align*}L=\mathcal{R}_\text{left}&=I \\ &= \int _0 ^1 \ln \left(\frac{1+x^2}{2} \right) \text{ d}x \\ &= \left[ x\ln \left(\frac{1+x^2}{2} \right) \right]_0 ^1 -\int_0 ^1 x \times \frac{x}{\frac{1+x^2}{2}}\text{ d}x \quad (\text{using integration by parts}) \\ &= \ln \left(\frac{2}{2} \right)-0 - 2 \int _0 ^1 \frac{x^2}{1+x^2}\text{ d}x \\ &= -2 \int _0 ^1 \left(1-\frac{1}{1+x^2} \right)\text{ d}x \\ &= -2\Big{[}x-\tan^{-1} x\Big{]}_0 ^1 \\ &= -2 \left(1-\tan^{-1}1 \right) \\ &= 2\left(\frac{\pi}{4}-1 \right) \\ &= \frac{\pi}{2}-2. \end{align*}$

HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

juantheron

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omegadot

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InteGrand

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