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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.

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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