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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 58 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 58 Maths Textbook Solution.

Answers (1)

Answer: \frac{e^{4}-2e^{2}}{4}

Hint: Use indefinite formula then put the limit to solve this integral

Given: \int_{1}^{2}e^{2x}\left ( \frac{1}{x}-\frac{1}{2x^{2}} \right )dx

Solution:

\begin{aligned} &\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x=2 \int_{1}^{2} e^{2 x}\left(\frac{1}{2 x}-\frac{1}{2 \times 2 x^{2}}\right) d x \\ &\text { Put } 2 x=t \Rightarrow 2 d x=d t \Rightarrow d x=\frac{d t}{2} \end{aligned}

When x=1 then t=2 and

When x=2  then t=4

Then \int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x=2 \int_{1}^{2} e^{2 x}\left(\frac{1}{2 x}-\frac{1}{4 x^{2}}\right) d x

\begin{aligned} &=2 \int_{2}^{4} e^{t}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) \frac{d t}{2} \\ &=\int_{2}^{4} e^{t}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) d t \\ &=\int_{2}^{4} e^{t} \frac{1}{t} d t-\int_{2}^{4} e^{t} \frac{1}{t^{2}} d t \end{aligned}

Apply integration by parts in 1st integral, then

\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x=\left[\frac{1}{t} \int e^{t} d t\right]_{2}^{4}-\int_{2}^{4}\left(\frac{d}{d t}\left(\frac{1}{t}\right) \int e^{t} d t\right) d t-\int_{2}^{4} e^{t} \frac{1}{t^{2}} d t

=\left[\frac{1}{t} e^{t}\right]_{2}^{4}-\int_{2}^{4}\left(\frac{-1}{t^{2}} e^{t}\right) d t-\int_{2}^{4} e^{t} \frac{1}{t^{2}} d t                            \quad\left[\begin{array}{l} \frac{d}{d x}\left(\frac{1}{x}\right)=\frac{d\left(x^{-1}\right)}{d x}=-1 x^{-1-1}=-1 \cdot x^{-2}=\frac{-1}{x^{2}} \\ \int e^{x} d x=e^{x} \end{array}\right] \\

=\left[\frac{1}{4} e^{4}-\frac{1}{2} e^{2}\right]+\int_{2}^{4} \frac{1}{t^{2}} e^{t} d t-\int_{2}^{4} \frac{1}{t^{2}} e^{t} d t

=\frac{1}{4}\left(e^{4}-2 e^{2}\right)=\frac{e^{4}-2 e^{2}}{4}

Posted by

infoexpert21

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