Question

If $g\left(x\right)=\underset{1}{\overset{x}{\int }}{e}^{t^2}$ $dt$ then the value of $h\left(x\right)=\underset{3}{\overset{x^3}{\int }}{e}^{t^2}dt$ equals

• A. $g\left(x^3\right)-g\left(3\right)$
• B. $g\left(x^3\right)+g\left(3\right)$
• C. $g\left(x^3\right)-3$
• D. $g\left(x^3\right)-3g\left(x\right)$

Answer 1

The correct option is A $g\left(x^3\right)-g\left(3\right)$

$\underset{3}{\overset{x^3}{\int }}{e}^{t^2}dt=\underset{3}{\overset{1}{\int }}{e}^{t^2}dt+\underset{1}{\overset{x^3}{\int }}{e}^{t^2}dt$
$\underset{3}{\overset{x^3}{\int }}{e}^{t^2}dt=\underset{1}{\overset{x^3}{\int }}{e}^{t^2}dt-\underset{1}{\overset{3}{\int }}{e}^{t^2}dt=g\left(x^3\right)-g\left(3\right)$