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Calculus / Calculus: Early Transcendentals 1 / Chapter 7.7 / Problem 81AE

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Improper integrals Evaluate the following improper

Chapter 7, Problem 81AE

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QUESTION:

Improper integrals Evaluate the following improper integrals (Putnam Exam, 1939).

a. $\int_{1}^{3} \frac{d x}{\sqrt{(x-1)(3-x)}}$

b. $\int_{1}^{\infty} \frac{d x}{e^{x+1}+e^{3-x}}$

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QUESTION:

Improper integrals Evaluate the following improper integrals (Putnam Exam, 1939).

a. $\int_{1}^{3} \frac{d x}{\sqrt{(x-1)(3-x)}}$

b. $\int_{1}^{\infty} \frac{d x}{e^{x+1}+e^{3-x}}$

ANSWER:

Solution

Step 1:

(a)We have to evaluate the given improper integral

$\int\limits_{1}^{3}\frac{dx}{\sqrt{(x-1)(3-x)}}$

We can write

$(x-1)(3-x)=-(x-2{)}^{2}+1$

Therefore

$\int\limits_{1}^{3}\frac{dx}{\sqrt{(x-1)(3-x)}}=\int\limits_{1}^{3}\frac{dx}{\sqrt{-(x-2{)}^{2}+1}}$

Now substitute

$x-2\ =u\Rightarrow{}dx=du$

$\Rightarrow{}$ $\int\limits_{\ }^{\ }\frac{dx}{\sqrt{-(x-2{)}^{2}+1}}=\int\limits_{\ }^{\ }\frac{du}{\sqrt{1-{u}^{2}}}$

$=sin{\ }^{-1}u$

Substitute back u=x-2

$\Rightarrow{}$ $\int\limits_{\ }^{\ }\frac{dx}{\sqrt{-(x-2{)}^{2}+1}}=sin{\ }^{-1}\ (x-2)=-sin{\ }^{-1}(2-x)$

$\int\limits_{1}^{3}\frac{dx}{\sqrt{(x-1)(3-x)}}=[-sin{\ }^{-1}\ (2-x){]}_{1}^{3}$

$=(\frac{\pi{}}{2}-(-\frac{\pi{}}{2}))$

$\ \ \ \ =\pi{}$

(since $\lim\limits_{{x} \rightarrow {{1}^{+}}}$ $(-sin{\ }^{-1}\ (2-x))=-$ $\frac{\pi{}}{2}$ and $\lim\limits_{{x} \rightarrow {{3}^{-}}}$ $(-sin{\ }^{-1}\ (2-x))=$ $\frac{\pi{}}{2}$ )

Hence

$\int\limits_{1}^{3}\frac{dx}{\sqrt{(x-1)(3-x)}}=\pi{}$

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Improper integrals Evaluate the following improper

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Solution Found!

Improper integrals Evaluate the following improper

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Chapter: 4 Problem: 10P

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