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Q. 40

Expert-verifiedFound in: Page 373

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals.

Use a graph to check your answer.

${\int}_{-3}^{2}\u200a\frac{1}{(x+5{)}^{2}}dx$

Ans: The exact value is, $\phantom{\rule{0ex}{0ex}}{\int}_{-3}^{2}\u200a\frac{1}{(x+5{)}^{2}}dx=\frac{5}{14}$

given expression,

${\int}_{-3}^{2}\u200a\frac{1}{(x+5{)}^{2}}dx$

The exact value is calculated as shown below,

$\begin{array}{r}{\int}_{-3}^{2}\u200a\frac{1}{(x+5{)}^{2}}dx\\ ={\int}_{-3}^{2}\u200a(x+5{)}^{-2}dx\\ ={\left[\frac{(x+5{)}^{-2+1}}{-2+1}\right]}_{-3}^{2}\\ ={\left[\frac{(x+5{)}^{-1}}{-1}\right]}_{-3}^{2}\\ ={\left[\frac{-1}{(x+5)}\right]}_{-3}^{2}\\ =\frac{-1}{(2+5)}+\frac{1}{(-3+5)}\\ =\frac{-1}{7}+\frac{1}{2}\\ =\frac{-2+7}{14}\\ =\frac{5}{14}\end{array}$

Therefore, the exact value is $\frac{5}{14}$.$\frac{5}{14}$

The required graph is,

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