Question

${\int }_{-4}^{-5}{e}^{{(x+5)}^2}dx+{\int }_{1/3}^{2/3}{e}^{{9(x-\frac{2}{3})}^2}dx$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. $-2$

Answer 1

Answer: (C)

$0$

We have given from the question

$\begin{array}{c}{\int }_{-4}^{-5}{e}^{{\left(x+5\right)}^2}dx+{\int }_{\frac{1}{3}}^{\frac{2}{3}}{e}^{9{\left(x-\frac{2}{3}\right)}^2}dx\\ Let\\ x+5=t\\ Now,\\ {\int }_1^0{e}^{t^2}dx+3{\int }_{\frac{1}{3}}^{\frac{2}{3}}{e}^{{\left(3x-2\right)}^2}dx\\ Again\\ Let3x-2=v\\ \therefore \frac{dv}{dx}=3\\ Now\\ {\int }_1^0{e}^{t^2}dx+\frac{1}{3}\times 3{\int }_{-1}^0{e}^{v^2}dv\\ {\int }_1^0{e}^{t^2}dx+{\int }_{-1}^0{e}^{v^2}dv\left[u=-x\right]\\ {\int }_1^0{e}^{t^2}dx+{\int }_1^0{e}^{v^2}dv=0\end{array}$

Hence, the option $\left(C\right)$ is the correct answer.