Question:
The value of the integral $ \int^{e^2}_{e^{-1}} \bigg | \frac{ \log_e \, x }{ x } \bigg | \, dx $ is
The value of the integral $ \int^{e^2}_{e^{-1}} \bigg | \frac{ \log_e \, x }{ x } \bigg | \, dx $ is
Updated On: Jun 14, 2022
- $\frac{3}{2} $
- $\frac{5}{2} $
- 3
- 5
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The Correct Option is B
Solution and Explanation
$ \int^{e^2}_{e^{-1}} \bigg | \frac{ \log_e \, x }{ x } \bigg | \, dx \, = \int^{1}_{e^{-1}} \bigg | \frac{ \log_e \, x }{ x } \bigg | \, dx \, - \int^{e^2}_{{1}} \bigg | \frac{ log_e \, x }{ x } \bigg | \, dx \, $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Bigg [$ since , 1 is turning point for $| \frac{ \log _e \, x }{x}| $ for $ + ve $ and $\, - ve $ values $ \Bigg ]$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = - \int^1_{e^{-1} }\frac{ \log _e \, x }{x} \, dx + \int^{e^2}_1 | \frac{ \log _e \, x }{x}| \, dx $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = -\frac{1}{2} [ ( \log_e \, x )^2 ]^1_{e^{-1}} + \frac{ 1 }{2} [ ( \log _ e \, x )^2]^{e^2 } _ 1 $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = - \frac{1}{2} \big \{ 0 - ( - 1 )^2 \big \} + \frac{1}{2} ( 2^2 - 0 ) = \frac{5}{2} $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Bigg [$ since , 1 is turning point for $| \frac{ \log _e \, x }{x}| $ for $ + ve $ and $\, - ve $ values $ \Bigg ]$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = - \int^1_{e^{-1} }\frac{ \log _e \, x }{x} \, dx + \int^{e^2}_1 | \frac{ \log _e \, x }{x}| \, dx $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = -\frac{1}{2} [ ( \log_e \, x )^2 ]^1_{e^{-1}} + \frac{ 1 }{2} [ ( \log _ e \, x )^2]^{e^2 } _ 1 $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = - \frac{1}{2} \big \{ 0 - ( - 1 )^2 \big \} + \frac{1}{2} ( 2^2 - 0 ) = \frac{5}{2} $
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
