Question

Evaluate the following using properties of integration

$\int_{-100}^{100} \: log( \sqrt{ {x}^{2} + 1} + x) \: dx$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\,I=\int^{100}_{-100}\log\left(\sqrt{{x}^{2}+1}+x\right)\,dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)}$

We know,

$\boxed{\bf{\displaystyle\int^{b}_{a}f(x)\,dx=\int^{b}_{a}f(a+b-x)\,dx}}$

So,

$\tt{\displaystyle\implies\,I=\int^{100}_{-100}\log\left(\sqrt{\left(100-100-x\right)^{2}+1}+\left(100-100-x\right)\right)\,dx}$

$\tt{\displaystyle\implies\,I=\int^{100}_{-100}\log\left(\sqrt{{x}^{2}+1}-x\right)\,dx}$

$\tt{\displaystyle\implies\,I=\int^{100}_{-100}\log\left\{\dfrac{\left(\sqrt{{x}^{2}+1}-x\right)\left(\sqrt{{x}^{2}+1}+x\right)}{\sqrt{{x}^{2}+1}+x}\right\}\,dx}$

$\tt{\displaystyle\implies\,I=\int^{100}_{-100}\log\left(\dfrac{{x}^{2}+1-{x}^{2}}{\sqrt{{x}^{2}+1}+x}\right)\,dx}$

$\tt{\displaystyle\implies\,I=\int^{100}_{-100}\log\left(\dfrac{1}{\sqrt{{x}^{2}+1}+x}\right)\,dx}$

$\tt{\displaystyle\implies\,I=-\int^{100}_{-100}\log\left(\sqrt{{x}^{2}+1}+x\right)\,dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)}$

Adding (1) and (2),

$\tt{\displaystyle\implies\,2I=\int^{100}_{-100}\log\left(\sqrt{{x}^{2}+1}+x\right)\,dx-\int^{100}_{-100}\log\left(\sqrt{{x}^{2}+1}+x\right)\,dx}$

$\tt{\displaystyle\implies\,2I=0}$

$\tt{\displaystyle\implies\,I=0}$

Answer 2

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

Given integral is

$\rm \: \displaystyle\int_{-100}^{100} \: log( \sqrt{ {x}^{2} + 1} + x) \: dx$

We know,

$\boxed{ \rm{ \:\begin{gathered}\begin{gathered}\bf\: \displaystyle\int_{-a}^{a}f(x)dx = \:\begin{cases} &\sf{0, \: \: if \: f( - x) = - f(x)} \\ \\ &\sf{2\displaystyle\int_{0}^{a}f(x)dx, \: \: if \: f( - x) = f(x)} \end{cases}\end{gathered}\end{gathered}}}$

Now, Let assume that

$\rm \: f(x) = log( \sqrt{ {x}^{2} + 1} + x)$

So,

$\rm \: f( - x) = log( \sqrt{ {x}^{2} + 1} - x)$

$\rm \:= log\bigg( \sqrt{ {x}^{2} + 1} - x \times \dfrac{ \sqrt{ {x}^{2} + 1} + x}{ \sqrt{ {x}^{2} + 1 } + x} \bigg)$

$\rm \:= log\bigg(\dfrac{ (\sqrt{ {x}^{2} + 1} )^{2} - {x}^{2} }{ \sqrt{ {x}^{2} + 1 } + x} \bigg)$

$\rm \:= log\bigg(\dfrac{ {x}^{2} + 1 - {x}^{2} }{ \sqrt{ {x}^{2} + 1 } + x} \bigg)$

$\rm \:= log\bigg(\dfrac{1}{ \sqrt{ {x}^{2} + 1 } + x} \bigg)$

$\rm \:= log( \sqrt{ {x}^{2} + 1 } + x)^{ - 1}$

$\rm \:= \: - \: log( \sqrt{ {x}^{2} + 1 } + x)$

$\rm \:= \: - \: f(x)$

So, we get

$\rm\implies \:\rm \:f(-x) \: = \: - \: f(x)$

Hence, Using the property, we get

$\rm\implies \:\displaystyle\int_{-100}^{100} \:log( \sqrt{ {x}^{2} + 1} + x)dx \: = \: 0$

$\rule{190pt}{2pt}$

Additional Information :-

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

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

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

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

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

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