Definite Integration Question - None

Question

Let $f, g:(0, \infty) \rightarrow R$ be two functions defined by $f(x)=\int_{-x}^{x}\left(|t|-t^{2}\right) e^{-t^{2}} d t$ and $g(x)=\int_{0}^{x^{2}} t^{1 / 2} e^{-t} d t$. Then the value of $\left(f\left(\sqrt{\log _{e} 9}\right)+g\left(\sqrt{\log _{e} 9}\right)\right)$is equal to

Accepted Answer

Given
$f(x)=\int_{-x}^{x}\left(|t|-t^{2}\right) e^{-t^{2}} d t$
$\Rightarrow f^{\prime}(x)=2 \cdot\left(|x|-x^{2}\right) e^{-x^{2}}$ ...(i)
$g(x)=\int_{0}^{x^{2}} t^{\frac{1}{2}} e^{-t} d t$
$g^{\prime}(x)=x e^{-x^{2}}(2 x)-0$
$f^{\prime}(x)+g^{\prime}(x)$
$=2 x^{-x^{2}}-2 x^{2} e^{-x^{2}}+2 x^{2} e^{-x^{2}}$
Integrating both sides w.r.t.x
$f(x)+g(x)=\int_{0}^{\alpha} 2 x e^{-x^{2}} d x$
Let $x^{2}={t}$
$\Rightarrow \int_{0}^{\sqrt{\alpha}} e^{-t} d t=\left[-e^{-t}\right]_{0}^{\sqrt{\alpha}}$
$=-{e}^{\left(\log _{0}(9)^{-1}\right)+1}$
Therefore,
$9(f(x)+g(x))=\left(1-\frac{1}{9}\right) 9=8$

Question:

Step-by-Step Solution

Master Definite Integration with Mathbee