Let f, g : (0, ∞) → R be two functions defined by f(x) = ∫(|t| − t^2)e^−t^2 dt; t ∈ [−x, x]

Question

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

(1) 6

(2) 9

(3) 8

(4) 10

Answer 1

Correct option is (3) 8

\(f(x)=\int\limits_{-x}^x|t|-t^2 e^{-t^2} d t\)

\(g(x)=\int\limits_0^{x^2} t^{\frac{1}{2}} e^{-t} d t \)

\(f(x)=2 \int\limits_0^x\left(t-t^2\right) e^{-t^2} d t\)

\( f(x)=2\left[\int\limits_0^x t^{-t^2} d t-\int\limits_0^x t^2 e^{-t^2} d t\right]\)

\(g(x)=\int\limits_0^{x^2} \sqrt{t} e^{-t} d t \quad \sqrt{t}=y \Rightarrow \frac{d t}{2 \sqrt{t}}=d y \)

\(g(x)=2 \int\limits_0^x y^2 \cdot e^{-y^2} d y \)

\(f(x)+g(x)=2\left(\frac{1-e^{-x^2}}{2}\right) \)

\(=1-e^{-x^2} \)

\(\Rightarrow 9\left(f\left(\sqrt{\log _e 9}\right)+g\left(\sqrt{\log _e 9}\right)\right)=9 \times\left(1-\frac{1}{9}\right)\)

\(=9 \times \frac{8}{9}=8\)