Let f, g:(0, infty) rightarrow mathbb{R} be t | vedclass

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(C) Given $f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} dt$. Since the integrand is an even function,$f(x) = 2 \int_{0}^{x} (t - t^2) e^{-t^2} dt$.By Lei

Vedclass · Accepted Answer

$8$

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(C) Given $f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} dt$. Since the integrand is an even function,$f(x) = 2 \int_{0}^{x} (t - t^2) e^{-t^2} dt$.By Leibniz's rule,$f'(x) = 2(x - x^2) e^{-x^2} = 2xe^{-x^2} - 2x^2e^{-x^2}$.Given $g(x) = \int_{0}^{x^2} t^{1/2} e^{-t} dt$. Let $t = u^2$,then $dt = 2u du$. When $t=0, u=0$ and when $t=x^2, u=x$.So,$g(x) = \int_{0}^{x} u e^{-u^2} (2u) du = 2 \int_{0}^{x} u^2 e^{-u^2} du$.By Leibniz's rule,$g'(x) = 2x^2 e^{-x^2}$.Now,$f'(x) + g'(x) = (2xe^{-x^2} - 2x^2e^{-x^2}) + 2x^2e^{-x^2} = 2xe^{-x^2}$.Integrating both sides with respect to $x$,$f(x) + g(x) = \int 2xe^{-x^2} dx = -e^{-x^2} + C$.Since $f(0) = 0$ and $g(0) = 0$,we have $f(0) + g(0) = 0$,so $C = 1$.Thus,$f(x) + g(x) = 1 - e^{-x^2}$.For $x = \sqrt{\log_{e} 9}$,$x^2 = \log_{e} 9$.Then $f(\sqrt{\log_{e} 9}) + g(\sqrt{\log_{e} 9}) = 1 - e^{-\log_{e} 9} = 1 - \frac{1}{9} = \frac{8}{9}$.Note: The provided options seem to be scaled by a factor of $9$. The value is $\frac{8}{9}$. If the question implies $9(f+g)$,the answer is $8$.

Let $f, g:(0, \infty) \rightarrow \mathbb{R}$ be two functions defined by $f(x)=\int_{-x}^x(|t|-t^2) e^{-t^2} dt$ and $g(x)=\int_0^{x^2} t^{1/2} e^{-t} dt$. Then the value of $(f(\sqrt{\log_{e} 9}) + g(\sqrt{\log_{e} 9}))$ is

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