Question:
The integral $ \int\limits^{1/2}_{-1/2} \bigg ([x] + \log \bigg ( \frac{1+ x}{1 - x} \bigg ) \bigg ) \, dx $ equals
The integral $ \int\limits^{1/2}_{-1/2} \bigg ([x] + \log \bigg ( \frac{1+ x}{1 - x} \bigg ) \bigg ) \, dx $ equals
Updated On: Jun 14, 2022
- $-\frac{1}{2}$
- 0
- 1
- $log \bigg ( \frac{1}{2} \bigg )$
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The Correct Option is A
Solution and Explanation
$ \int ^{1/2}_{-1/2} \bigg ([x] + log \bigg ( \frac{1+ x}{1 - x} \bigg ) \bigg ) \, dx \, $
$ = \int^{1/2}_{-1/2} [x] \, dx + \int^{1/2}_{-1/2} \, log \bigg (\frac{1+ x }{1 - x } \bigg ) dx $
$ = \int^{1/2}_{-1/2} [x] \, dx + 0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \bigg [\because \bigg ( \frac{ 1 + x }{ 1 - x} \bigg ) $ is an odd function $\bigg ]$
$ = \int^{0}_{-1/2} [x] \, dx + \int^{1/2}_{0} [x] \, dx = \int^{0}_{-1/2} ( - 1) \, dx + \int^{1/2}_{0} \, dx$
$=[ x]^0_{-1/2} = - \bigg ( 0 + \frac{1}{2} \bigg ) =- \frac{1}{2}$
$ = \int^{1/2}_{-1/2} [x] \, dx + \int^{1/2}_{-1/2} \, log \bigg (\frac{1+ x }{1 - x } \bigg ) dx $
$ = \int^{1/2}_{-1/2} [x] \, dx + 0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \bigg [\because \bigg ( \frac{ 1 + x }{ 1 - x} \bigg ) $ is an odd function $\bigg ]$
$ = \int^{0}_{-1/2} [x] \, dx + \int^{1/2}_{0} [x] \, dx = \int^{0}_{-1/2} ( - 1) \, dx + \int^{1/2}_{0} \, dx$
$=[ x]^0_{-1/2} = - \bigg ( 0 + \frac{1}{2} \bigg ) =- \frac{1}{2}$
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Concepts Used:
Integration by Parts
Integration by Parts is a mode of integrating 2 functions, when they multiplied with each other. For two functions ‘u’ and ‘v’, the formula is as follows:
∫u v dx = u∫v dx −∫u' (∫v dx) dx
- u is the first function u(x)
- v is the second function v(x)
- u' is the derivative of the function u(x)
The first function ‘u’ is used in the following order (ILATE):
- 'I' : Inverse Trigonometric Functions
- ‘L’ : Logarithmic Functions
- ‘A’ : Algebraic Functions
- ‘T’ : Trigonometric Functions
- ‘E’ : Exponential Functions
The rule as a diagram:
