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GIF Integral ∫[log_π x] d(logₑx) from e to π² | JEE

JEE Maths question with a full step-by-step solution.

Question

The value of

\displaystyle\int_{e}^{\pi^{2}}\big[\log_{\pi}x\big]\,d(\log_{e}x)

, where

[\,\cdot\,]

denotes the greatest integer function, is

2\log_{e}\pi

\log_{e}\pi

correct

1

0

Solution

Step 1: Substitute to make the variable of integration explicit. Put

\log_{e}x=t

, so that

d(\log_{e}x)=dt

. When

x=e

t=1

; when

x=\pi^{2}

t=\log_{e}\pi^{2}=2\log_{e}\pi

. Also

x=e^{t}

, hence

\log_{\pi}x=\log_{\pi}e^{t}=t\,\log_{\pi}e.

Step 2: The integral becomes

I=\int_{1}^{2\log_{e}\pi}\big[t\,\log_{\pi}e\big]\,dt.

Now

\log_{\pi}e=\dfrac{1}{\log_{e}\pi}

, so

t\log_{\pi}e=\dfrac{t}{\log_{e}\pi}

, and

t

runs from

1

2\log_{e}\pi

. Therefore

t\log_{\pi}e

runs from

\dfrac{1}{\log_{e}\pi}

up to

2

. Step 3: Locate where the greatest integer value changes. We have

\big[t\log_{\pi}e\big]=0

while

0\le t\log_{\pi}e<1

, i.e. for

t<\log_{e}\pi

, and

\big[t\log_{\pi}e\big]=1

while

1\le t\log_{\pi}e<2

, i.e. for

\log_{e}\pi\le t<2\log_{e}\pi

. Splitting at

t=\log_{e}\pi

I=\int_{1}^{\log_{e}\pi}0\,dt+\int_{\log_{e}\pi}^{2\log_{e}\pi}1\,dt.

Step 4: Evaluate:

I=0+\big(2\log_{e}\pi-\log_{e}\pi\big)=\log_{e}\pi.

Correct answer: (2)

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