Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Given : ${\int }_{-4}^{-5}{e}^{(x+5{)}^2}dx+3{\int }_{\frac{1}{3}}^{\frac{2}{3}}{e}^{9{(x-\frac{2}{3})}^2}$

$I={\int }_{-4}^{-5}{e}^{(x+5{)}^2}dx+3{\int }_{\frac{1}{3}}^{\frac{2}{3}}{e}^{9{(x-\frac{2}{3})}^2}{I}_1={\int }_{-4}^{-5}{e}^{(x+5{)}^2}dx$

Let : $x+5=tx\to -4t\to 1$

$dx=dtx\to -5t\to 0I_1={\int }_1^0{e}^{t^2}dt{I}_2=3{\int }_{\frac{1}{3}}^{\frac{2}{3}}{e}^{9{(x-\frac{2}{3})}^2}$

Let : $3x-2=tx\to \frac{1}{3}t\to -1$

$3dx=dtx\to \frac{2}{3}t\to 0I_2={\int }_{-1}^0{e}^{t^2}dtI=I_1+I_2={\int }_1^0{e}^{t^2}dt+{\int }_{-1}^0{e}^{t^2}dt=0$

Hence the correct answer is $0$