Question

The value of the definite integral $\underset{-1}{\overset{1}{\int }}{e}^{-x^4}\left(1+\ln \left(x+\sqrt{x^2+1}\right)+5x^3-4x^4\right)dx$ is equal to

• A. $4e$
• B. $\frac{4}{e}$
• C. $2e$
• D. $\frac{2}{e}$

Answer 1

Answer: (D)

As, $f(x)=1+\ln \left(x+\sqrt{x^2+1}\right)+5x^3-4x^4$
Clearly, $\ln \left(\sqrt{x^2+1}+x\right),5x^3$ are odd functions, so
$I=\underset{-1}{\overset{1}{\int }}\left(1-4x^4\right)e^{-x^4}dx=2\underset{0}{\overset{1}{\int }}\left(1-4x^4\right)e^{-x^4}dx=2\underset{0}{\overset{1}{\int }}\frac{d}{dx}\left(x\cdot e^{-x^4}\right)dx=2{\left(\frac{x}{e^{x^4}}\right)}_0^1=2\left(\frac{1}{e}-0\right)=\frac{2}{e}$