If f(x)={| x|}{x²}, then ₁/e}^{e} f(x) \, dx=

Correct Answer is : 2(1−1e)2≤ft(1-1/e)2(1−e1)

Correct Answer is : 2(1−1e)2≤ft(1-1/e)2(1−e1)

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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

f(x)=\frac{|\log x|}{x^{2}},

\int\limits_{\frac{1}{e}}^{e} f(x) \, dx=

Solution:

Consider the function.

f(x)=\frac{|\log x|}{x^{2}}

The function can be written as,

f(x)=\begin{cases} \frac{-\log x}{x^{2}}, & 0 < x<1 \\ \frac{\log x}{x^{2}}, & 1 < x < \infty \end{cases}

Therefore,

\int\limits_{\frac{1}{e}}^{e} f(x) \, dx=\int\limits_{\frac{1}{e}}^{1} f(x) \, dx+\int\limits_{1}^{e} f(x) \, dx

=\int\limits_{\frac{1}{e}}^{1} \frac{-\log x}{x^{2}} \, dx+\int\limits_{1}^{e} \frac{\log x}{x^{2}} \, dx

=-\left[ \frac{\log x}{x}-\frac{1}{x} \right]_{\frac{1}{e}}^{1}+\left[ \frac{-\log x}{x}-\frac{1}{x} \right]_{1}^{e}

=(1+e-e)-\left( \frac{1}{e}+\frac{1}{e}-1 \right)

Solve further,

\int\limits_{\frac{1}{e}}^{e} f(x) \, dx=1-\frac{2}{e}+1

=2-\frac{2}{e}

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

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