Subject: Mathematics — Definite Integration
Question: Let
$\psi_{1}:\lbrack 0,\infty\rbrack\mathbb{\rightarrow R,}\psi_{2}:\lbrack 0,\infty\rbrack\mathbb{\rightarrow R,}f:\lbrack 0,\infty\rbrack\mathbb{\rightarrow R}$
and $g:\lbrack 0,\infty\rbrack\mathbb{\rightarrow R,}$ be functions such
that $f(0) = f(0) = 0$,
$\psi_{1}(x) = e^{- x} + x,\ x \geq 0$,
$\psi_{2}(x) = x^{2} - 2x - 2e^{- x} + 2,\ x \geq 0$,
$f(x) = \int_{- x}^{x}\mspace{2mu}\left( |t| - t^{2} \right)e^{- t^{2}}dt,x > 0$
and
$g(x) = \int_{0}^{x^{2}}\mspace{2mu}\mspace{2mu}\sqrt{t}{\ e}^{- t}dt,x > 0$
Which of the following statements is TRUE? then find the value of
$f(\sqrt{ln3}) + g(\sqrt{ln3})$
Answer: $\frac{2}{3}$
Solution: $\because f(x) = \int_{- x}^{x}\mspace{2mu}\left( |t| - t^{2} \right)e^{- t^{2}}dt,x > 0$\
$$\therefore\ f(x) = 2\int_{0}^{x}\mspace{2mu}\mspace{2mu}\left( t - t^{2} \right)e^{- t^{2}}dt
$$$f^{'}(x) = 2\left( x - x^{2} \right)e^{- x^{2}}$ .....(i)
[IMAGE] Hence option (d) is wrong.\
$g^{'}(x) = xe^{- x^{2}}2x$ .....(ii)\
Add (i) and (ii),\
$${f^{'}(x) + g^{'}(x) = 2xe^{- x^{2}}
}{f(x) + g(x) = - e^{- x^{2}} + c
}{\because\ f(0) = g(0) = 0
}{f(x) + g(x) = - e^{- x^{2}} + 1
}{f(\sqrt{\mathcal{l}ln3}) + g(\sqrt{\mathcal{l}ln3}) = 1 - \frac{1}{3} = \frac{2}{3}}$$