Question

Integrate the following

$\int _{ - 100}^{100} ( \frac{ {x}^{2} }{1 + {2}^{x} } ) \: dx$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int_{ - 100}^{100}\rm \frac{ {x}^{2} }{1 + {2}^{x} } \: dx$

Let assume that

$\rm :\longmapsto\:I = \displaystyle\int_{ - 100}^{100}\rm \frac{ {x}^{2} }{1 + {2}^{x} } \: dx - - - (1)$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:I = \displaystyle\int_{ - 100}^{100}\rm \frac{ {( - x)}^{2} }{1 + {2}^{ - x} } \: dx$

$\rm :\longmapsto\:I = \displaystyle\int_{ - 100}^{100}\rm \frac{ {x}^{2} }{1 + \dfrac{1}{ {2}^{x} } } \: dx$

$\rm :\longmapsto\:I = \displaystyle\int_{ - 100}^{100}\rm \frac{ {x}^{2} {2}^{x} }{{2}^{x} + 1} \: dx - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2I = \displaystyle\int_{ - 100}^{100}\rm \frac{ {x}^{2} + {x}^{2} {2}^{x} }{{2}^{x} + 1} \: dx$

$\rm :\longmapsto\:2I = \displaystyle\int_{ - 100}^{100}\rm \frac{ {x}^{2}(1 + {2}^{x})}{{2}^{x} + 1} \: dx$

$\rm :\longmapsto\:2I = \displaystyle\int_{ - 100}^{100}\rm {x}^{2} \: dx$

$\rm :\longmapsto\:2I = \bigg[\dfrac{ {x}^{2 + 1} }{2 + 1} \bigg]_{ - 100}^{100}$

$\rm :\longmapsto\:2I = \bigg[\dfrac{ {x}^{3} }{3} \bigg]_{ - 100}^{100}$

$\rm :\longmapsto\:2I = \dfrac{1}{3}[ {(100)}^{3} - {( - 100)}^{3}]$

$\rm :\longmapsto\:2I = \dfrac{1}{3}[ 1000000 + 1000000]$

$\rm :\longmapsto\:2I = \dfrac{1}{3}[ 2000000]$

$\rm :\longmapsto\:I = \dfrac{1000000}{3}$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle\int_{ - 100}^{100}\rm \frac{ {x}^{2} }{1 + {2}^{x} } \: dx = \frac{1000000}{3}}}$

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MORE TO KNOW

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

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

$\boxed{\tt{ \displaystyle\int_0^a\rm f(x) \: dx \: = \: \displaystyle\int_0^a\rm f(a - x)dx}}$

$\boxed{\tt{ \displaystyle\int_{ - a}^a\rm f(x) \: dx \: = \:2 \displaystyle\int_0^a\rm f( x)dx \: \: if \: f( - x) = f(x)}}$

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

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

$\boxed{\tt{ \displaystyle\int_{0}^{2a}\rm f(x) \: dx \: = \displaystyle\int_0^a\rm f(x)dx\: \: if \: f(2a - x) = f(x)}}$