Math GRE - #17

Let $h$ be the function defined by $$h(x)=\int_0^{x^2}{e^{x+t}dt}.$$ Then $h'(1)=$

• $e-1$
• $e^2$
• $e^2-e$
• $2e^2$
• $3e^2-e$

Solution :

The answer is $3e^2-e$.

This can be solved with applications of the product rule, the chain rule, and the fundamental theorem of calculus.

First let us recall that the derivative of an integral is itself, that is: $$\frac{d}{dx}\int_a^x{f(x)}=f(x).$$
When the upper limit is a function, we must apply the chain rule: $$\frac{d}{dx}\int_a^{x^2}{f(x)}=2x\cdot f(x^2).$$
For this particular question, we have: $$h(x)=\int_0^{x^2}{e^{x+t}dt}=e^x\cdot\int_0^{x^2}{e^{t}dt}=e^x\cdot g(x)$$ for $g(x)=\int_0^{x^2}{e^{t}dt}$. The product rule tells us that $$h'(x)=e^x\cdot g(x)+e^x\cdot g'(x)=e^x\cdot\int_0^{x^2}{e^{t}dt}+e^x\cdot\left(2x\cdot e^{x^2}\right).$$
Upon substituting $1$ for $x$, we obtain the answer: $$h'(1)=e\cdot(e-1)+e\cdot(2\cdot e)=3e^2-e.$$