SATHEE: Definite Integration Question 18

Definite Integration Question 18

Question 18

For $x > 0$ , let $f (x) = \int_{1}^{x} \frac{\ln t}{1 + t} d t$ . Find the function $f (x) + f (1 / x)$ and show that $f (e) + f (1 / e) = 1 / 2$ , where $\ln t = \log_{e} t$ .

$(2000, 5 M)$

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Answer:

Correct Answer: 18. $\frac{1}{2} (\ln x)^{2}$

Solution:

$f (x) = \int_{1}^{x} \frac{\ln t}{1 + t} d t$ for $x > 0$

[given]

Now, $f (1 / x) = \int_{1}^{1 / x} \frac{\ln t}{1 + t} d t$

Put $t = 1 / u \Rightarrow d t = (- 1 / u^{2}) d u$

$\begin{aligned} ∴ f (1 / x) & = \int_{1}^{x} \frac{\ln (1 / u)}{1 + 1 / u} \cdot \frac{(- 1)}{u^{2}} d u & = \int_{1}^{x} \frac{\ln u}{u (u + 1)} d u = \int_{1}^{x} \frac{\ln t}{t (1 + t)} d t \end{aligned}$

Now, $f (x) + f \frac{1}{x} = \int_{1}^{x} \frac{\ln t}{(1 + t)} d t + \int_{1}^{x} \frac{\ln t}{t (1 + t)} d t$

$= \int_{1}^{x} \frac{(1 + t) \ln t}{t (1 + t)} d t = \int_{1}^{x} \frac{\ln t}{t} d t = \frac{1}{2} {[(\ln t)^{2}]}_{1}^{x} = \frac{1}{2} (\ln x)^{2}$

Put $x = e$ ,

$f (e) + f \frac{1}{e} = \frac{1}{2} (\ln e)^{2} = \frac{1}{2}$

Hence proved.

Definite Integration Question 18

Question 18

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Definite Integration Question 18

Question 18

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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