Question

Integration of -1 to 1 (x^2/1 +e^x) dx

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_{-1}^1 \rm \frac{ {x}^{2} }{1 + {e}^{x} } \: dx$

Let assume that

$\rm :\longmapsto\: I \: = \: \displaystyle\int_{-1}^1 \rm \frac{ {x}^{2} }{1 + {e}^{x} } \: dx \: - - - (1)$

We know that,

$\boxed{ \sf{ \:\displaystyle\int_{a}^b \rm f(x)dx = \displaystyle\int_{a}^b \rm f(a + b - x)dx}}$

Change x to - x, we get

$\rm :\longmapsto\: I \: = \: \displaystyle\int_{-1}^1 \rm \frac{ {( - x)}^{2} }{1 + {e}^{ - x} } \: dx \:$

$\rm :\longmapsto\: I \: = \: \displaystyle\int_{-1}^1 \rm \frac{ { x}^{2} }{1 + \dfrac{1}{ {e}^{x} } } \: dx \:$

$\rm :\longmapsto\: I \: = \: \displaystyle\int_{-1}^1 \rm \frac{ { x}^{2} }{\dfrac{ {e}^{x} + 1}{ {e}^{x} } } \: dx \:$

$\rm :\longmapsto\: I \: = \: \displaystyle\int_{-1}^1 \rm \frac{ {x}^{2} {e}^{x} }{1 + {e}^{x} } \: dx \: - - - (2)$

Adding equation (1) and (2), we get

$\rm :\longmapsto\: 2I \: = \: \displaystyle\int_{-1}^1 \rm \frac{ {x}^{2}}{1 + {e}^{x} } \: dx + \: \displaystyle\int_{-1}^1 \rm \frac{ {x}^{2} {e}^{x} }{1 + {e}^{x} } \: dx$

$\rm :\longmapsto\: 2I \: = \: \displaystyle\int_{-1}^1 \rm \frac{ {x}^{2}(1 +{e}^{x}) }{1 + {e}^{x} } \: dx$

$\rm :\longmapsto\: 2I \: = \: \displaystyle\int_{-1}^1 \rm {x}^{2} \: dx$

$\rm :\longmapsto\: 2I \: = \: \bigg(\dfrac{ {x}^{2 + 1} }{2 + 1} \bigg) _{-1}^1$

$\rm :\longmapsto\: 2I \: = \: \bigg(\dfrac{ {x}^{3} }{3} \bigg) _{-1}^1$

$\rm :\longmapsto\: 2I \: = \: \dfrac{ {1} }{3}( {1}^{3} - {( - 1)}^{3} )$

$\rm :\longmapsto\: 2I \: = \: \dfrac{ {1} }{3}(1 - ( - 1))$

$\rm :\longmapsto\: 2I \: = \: \dfrac{ {1} }{3}(1 + 1)$

$\rm :\longmapsto\: 2I \: = \: \dfrac{ {1} }{3}(2)$

$\bf :\longmapsto\: I \: = \: \dfrac{ {1} }{3}$

Additional Information :-

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = \: \displaystyle\int_{a}^b \rm f(y)dy}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = \: - \: \displaystyle\int_{b}^a \rm f(x)dx}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{0}^a \rm f(x)dx \: = \: \displaystyle\int_{0}^a \rm f(a - x)dx}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{ - a}^a \rm f(x)dx \: =2 \: \displaystyle\int_{0}^a \rm f(x)dx \: \: if \: f( - x) = f(x)}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{ - a}^a \rm f(x)dx \: =0 \: \: if \: f( - x) = - \: f(x)}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{0}^{2a} \rm f(x)dx \: =2 \: \displaystyle\int_{0}^a \rm f(x)dx \: \: if \: f(2a - x) = f(x)}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{0}^{2a} \rm f(x)dx \: =0 \: \: if \: f(2a - x) = - \: f(x)}}}$