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  4. The integral ∫ limits01 (1/.7[(1)x.]) dx, where [.] denotes the greatest integer function is equal to

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Q. The integral $\int\limits_{0}^{1} \frac{1}{\left.7^{\left[\frac{1}{x}\right.}\right]} dx$, where $[.]$ denotes the greatest integer function is equal to

JEE MainJEE Main 2022Integrals

Solution:

$\int\limits_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx =-\int\limits_{1}^{0} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx$

$=(-1)\left[\int\limits_{1}^{1 / 2} \frac{1}{7} dx +\int\limits_{1 / 2}^{1 / 3} \frac{1}{7^{2}} dx +\int\limits_{1 / 3}^{1 / 4} \frac{1}{7^{3}} dx +\ldots \ldots \infty\right] =\left(\frac{1}{7}+\frac{1}{2 \cdot 7^{2}}+\frac{1}{3 \cdot 7^{3}}+\ldots \infty\right)-\left(\frac{1}{7 \cdot 2}+\frac{1}{7^{2} \cdot 3}+\frac{1}{7^{2} \cdot 4} \ldots \infty\right)$
$=-\ln \left(1-\frac{1}{7}\right)-7\left(\frac{1}{7^{2} \cdot 2}+\frac{1}{7^{3} \cdot 3}+\frac{1}{7^{4} \cdot 4}+\ldots \ldots \infty\right)$
$={\left[\text{as} \ln (1+ x )= x -\frac{ x ^{2}}{2}+\frac{ x ^{3}}{3}-\frac{ x ^{4}}{4} \ldots \infty\right] }$
$\left[\right.$ as $\left.\ln (1-x)=-\left(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4} \ldots \infty\right)\right]$
$=6 \ln \frac{6}{7}-7\left(-\ln \left(1-\frac{1}{7}\right)-\frac{1}{7}\right)$
$=6 \ln \frac{6}{7}+1$