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Chapter 2: Problem 18

\begin{aligned} &\text { Find the value of 'a' such that }\\\ &\int_{0}^{a} \frac{d x}{e^{x}+4 e^{-x}+5}=\ln \sqrt[3]{2} . \end{aligned}

Short Answer

Expert verified
Answer: The value of 'a' is $$a = \ln(4 \sqrt[3]{2} - 4)$$

Step by step solution

01

Integrating the given expression

Integrate the given expression with respect to x: $$ \int \frac{1}{e^x + 4e^{-x} + 5} \, dx $$
02

Substitution Method

Substitute \(u = e^x\), and then find \(\frac{du}{dx}\): $$ \frac{du}{dx} = e^x \Rightarrow dx = \frac{du}{u} $$ Substitute \(e^x\) with \(u\) in the original expression and change \(dx\) to \(\frac{du}{u}\), we get: $$ \int \frac{1}{u + 4/u + 5} \frac{du}{u} $$
03

Simplify the integral

Simplify the integral and find the indefinite integral: $$ \int \frac{1}{u^2 + 5u + 4} du $$ Factor the denominator: $$ \int \frac{1}{(u + 4)(u + 1)} du $$
04

Apply Partial Fraction Decomposition

Use partial fraction decomposition for the expression: $$ \frac{1}{(u + 4)(u + 1)} = \frac{A}{u + 4} + \frac{B}{u + 1} $$ Solve for A and B: $$ A(u + 1) + B(u + 4) = 1 $$ By comparing coefficients, we find A and B: $$A = \frac{1}{3}, \quad B = -\frac{1}{3}$$ Now our integral becomes: $$ \int \left( \frac{1}{3(u + 4)} - \frac{1}{3(u + 1)} \right) du $$
05

Integrate with the substitution

Integrate the expression with respect to u: $$ \int \left( \frac{1}{3(u + 4)} - \frac{1}{3(u + 1)} \right) du = \frac{1}{3} \ln |u + 4| - \frac{1}{3} \ln |u + 1| + C $$ Substitute back \(u = e^x\): $$ \frac{1}{3} \ln |e^x + 4| - \frac{1}{3} \ln |e^x + 1| + C $$
06

Definite integral

Find the definite integral with the given limits from 0 to a: $$ \int_{0}^{a} \frac{dx}{e^x + 4e^{-x} + 5} = \left[\frac{1}{3} \ln |e^x + 4| - \frac{1}{3} \ln |e^x + 1| \right]_0^a $$
07

Solve for 'a'

Set the definite integral equal to the natural logarithm of the cube root of 2: $$ \left[\frac{1}{3} \ln |e^a + 4| - \frac{1}{3} \ln |e^a + 1| - \frac{1}{3} \ln |4| + \frac{1}{3} \ln |1| \right] = \ln \sqrt[3]{2} $$ Solve for 'a': $$ \frac{1}{3} \ln \frac{|e^a + 4|}{|e^a + 1|} - \frac{1}{3} \ln 4 = \ln \sqrt[3]{2} $$ Both sides are in natural logarithm form, therefore: $$ \frac{|e^a + 4|}{|e^a + 1|} = \frac{4 \sqrt[3]{2}}{4} $$ Solve for a: $$ e^a + 4 = 4 \sqrt[3]{2} $$ Subtract 4: $$ e^a = 4 \sqrt[3]{2} - 4 $$ Take natural logarithm of both sides: $$ a = \ln(4 \sqrt[3]{2} - 4) $$ So, the value of 'a' is: $$ a = \ln(4 \sqrt[3]{2} - 4) $$

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Most popular questions from this chapter

Prove that (i) \(\frac{99 \pi}{400}<\int_{1}^{100} \frac{\tan ^{-1} x}{x^{2}} d x<\frac{99 \pi}{200}\) (ii) \(\frac{609(\ln 2)^{2}}{4}<\int_{2}^{5} x^{3}(\ln x)^{2} d x<\frac{609(\ln 5)^{2}}{4}\) (iii) \(\left(1-\mathrm{e}^{-1}\right) \ln 10<\int_{1}^{10} \frac{1-\mathrm{e}^{-x}}{\mathrm{x}} \mathrm{dx}<\ln 10\) (iv) \(\frac{1}{10 \sqrt{2}} \leq \int_{0}^{1} \frac{x^{9}}{\sqrt{1+x}} d x \leq \frac{1}{10}\).

Prove that (i) \(\int_{0}^{1} \frac{x^{m-1}}{1+x^{n}} d x=\frac{1}{m}-\frac{1}{m+n}+\frac{1}{m+2 n}-\frac{1}{m+3 n}+\ldots\) (ii) \(\int_{0}^{x} \frac{\sin x}{x} d x=x-\frac{x^{3}}{3.3 !}+\frac{x^{5}}{5.5 !}-\ldots\) (iii) \(\int_{a}^{b} \frac{\mathrm{e}^{x}}{x} \mathrm{dx}=\ln \frac{\mathrm{b}}{\mathrm{a}}+(\mathrm{b}-\mathrm{a})+\frac{\mathrm{b}^{2}-\mathrm{a}^{2}}{2.2 !}+\frac{\mathrm{b}^{3}-\mathrm{a}^{3}}{3.3 !}+\ldots\) (iv) \(\int_{0}^{1} \frac{\tan ^{-1} x}{x} d x=\sum_{0}^{\infty}(-1)^{n} \frac{1}{(2 n+1)^{2}}\).

Prove that \(\int_{a}^{b} \frac{d x}{\sqrt{\\{(x-a)(b-x)\\}}}=\pi\), \(\int_{a}^{b} \frac{x d x}{\sqrt{\\{(x-a)(b-x)\\}}}=\frac{1}{2} \pi(a+b)\) (i) by means of the substitution \(\mathrm{x}=\mathrm{a}+(\mathrm{b}-\mathrm{a}) \mathrm{t}^{2}\), (ii) bymeans of the substitution \((\mathrm{b}-\mathrm{x})(\mathrm{x}-\mathrm{a})=\mathrm{t}\), and (iii) by means of the substitution \(x=a \cos ^{2} t\) \(+b \sin ^{2} t\)

Let p be a polynomial of degree atmost 4 such that \(\mathrm{p}(-1)=\mathrm{p}(1)=0\) and \(\mathrm{p}(0)=1\). If \(\mathrm{p}(\mathrm{x}) \leq 1\) for \(x \in[-1,1]\), find the largest value of \(\int^{1} p(x) d x\)

Estimate \(\int_{0}^{3} \mathrm{f}(\mathrm{x}) \mathrm{d} x\) if \(\mathrm{it}\) is known that \(f(0)=10, f(0.5)=13, f(1)=14, f(1.5)=16, f(2)=18\) \(\mathrm{f}(2.5)=10, \mathrm{f}(3)=6 \mathrm{by}\) (a) the trapezoidal method. (b) Simpson's method.
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