Question:
If $I_1=\int\limits_{e} ^{_e2}\frac{dx}{\log\,x}$ and $I_2=\int\limits_{1} ^{2}\frac{e^x}{x}dx$ , then
If $I_1=\int\limits_{e} ^{_e2}\frac{dx}{\log\,x}$ and $I_2=\int\limits_{1} ^{2}\frac{e^x}{x}dx$ , then
Updated On: Jul 6, 2022
- $I_1=I_2$
- $2I_1=I_2$
- $I_1=2I_2$
- none of these.
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The Correct Option is A
Solution and Explanation
$I=\int\limits_{e}^{e^2} \frac{dx}{log\,x}$ Put log $x=z$,
$\therefore x=e^{z} $
$\therefore dx=e^{z} dz$
When $x = e, z = log e = 1$
$x=e^{2}, z=log\,e^{2}=2 \,log e=2$
$\therefore I_{1}=\int\limits_{1}^{2} \frac{e^{z}dz}{z}=\int\limits_{1}^{2} \frac{e^{x}}{z} dx=I_{2}$
$\therefore I_{1}=I_{2}$
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
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Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
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- The problem of getting the area bounded by the graph of a function under given situations.