IIT JEE Problem

Calculus Level 4

T h e v a l u e o f ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x i s : The\quad value\quad of\quad \int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 }{ (1-x) }^{ 4 } }{ 1+{ x }^{ 2 } } } dx\quad is:

22 7 − π \frac { 22 }{ 7 } -\pi -1 0 71 15 − 3 2 π \frac { 71 }{ 15 } -\frac { 3 }{ 2 } \pi 2 105 . \frac { 2 }{ 105 } . 1

Vishnu C by vishnu c , 22, India

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1 solution

Tom Engelsman
Nov 12, 2016

The above integrand can be expanded and simplified into:

[ x 4 ∗ ( 1 − x ) 4 ] / ( 1 + x 2 ) = x 4 ∗ ( 1 − 4 x + 6 x 2 − 4 x 3 + x 4 ) / ( 1 + x 2 ) [x^4 * (1-x)^4]/(1 + x^2) = x^4 * (1 - 4x + 6x^2 - 4x^3 + x^4) / (1 + x^2) ;

or ( x 4 − 4 x 5 + 6 x 6 − 4 x 7 + x 8 ) / ( 1 + x 2 ) (x^4 - 4x^5 + 6x^6 - 4x^7 + x^8) / (1 + x^2) ;

or ( x 6 − 4 x 5 + 5 x 4 − 4 x 2 + 4 ) − 4 / ( 1 + x 2 ) (x^6 - 4x^5 + 5x^4 - 4x^2 + 4) - 4/(1 + x^2) (i).

Integrating (i) with respect to x gives: ∫ 0 1 x 6 − 4 x 5 + 5 x 4 − 4 x 2 + 4 − 4 / ( 1 + x 2 ) d x \int_0^1 x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - 4/(1 + x^2) \, dx ;

or ( 1 / 7 ) ∗ x 7 − ( 2 / 3 ) ∗ x 6 + x 5 − ( 4 / 3 ) ∗ x 3 + 4 x − 4 ∗ a r c t a n ( x ) (1/7)*x^7 - (2/3)*x^6 + x^5 - (4/3)*x^3 + 4x - 4*arctan(x) (ii)

which plugging in the limits of integration produces:

[ 1 / 7 − 2 / 3 + 1 − 4 / 3 + 4 − 4 ∗ a r c t a n ( 1 ) ] − [ 0 − 4 ∗ a r c t a n ( 0 ) ] [1/7 - 2/3 + 1 -4/3 + 4 - 4*arctan(1)] - [0 - 4*arctan(0)] ;

or ( 1 / 7 + 5 − 2 ) − 4 ∗ π / 4 (1/7 + 5 - 2) - 4*\pi/4 ;

or 22 / 7 − π 22/7 - \pi .

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