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Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 2 Maths textbook Solution.

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Answer: log 2

Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer.

Given: \int_{-2}^{3}\frac{1}{x+7}dx

Solution:\int_{-2}^{3}\frac{1}{x+7}dx

Putting x+7=t\Rightarrow dx=dt

And when x=-2 then t=-2+7=5

x=3 then t=3+7=10

Then,\int_{-2}^{3}\frac{1}{x+7}dx=\int_{5}^{10}\frac{1}{t}dt=\left [ log|t| \right ]^{10}_{5}                                                                    \left [ \int \frac{1}{x}dx-log|x| \right ]

=\left [ log|10|-log|5| \right ]                                                                                                    \left [ log\: a-log\: b=log\frac{a}{b} \right ]

=log10-log5
=log\frac{10}{5}

 

=log 2

 

Posted by

infoexpert21

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