Question

If $g\left(x\right)={\int }_1^x{e}^{t^2}dt$ then the value of ${\int }_3^{x^3}{e}^{t^2}dt$ equals

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

Answer 1

Answer: (A)

$g\left(x^3\right)-g\left(3\right)$

$g\left(x\right)={\int }_1^x{e}^{t^2}dt⟹g^{\prime }\left(x\right)=e^{x^2}⟹g\left(x\right)=\int e^{t^2}dt⟹{\int }_3^{x^3}{e}^{t^2}dt=g\left(x^3\right)-g\left(3\right)$