# Calculating $H'(x)$ given $H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$

Given $$\displaystyle H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$$, we want to find $$H'(x)$$.

First, we rewrite $$H(x)$$ as follows:

\begin{align} &= \int_0^{x^2 + 2x} e^{-t^2} dt + \int_{x^3 + 1}^0 e^{-t^2} dt \qquad &\text{Properties of integrals} \\ &= \int_0^{x^2 + 2x} e^{-t^2} dt - \int_{0}^{x^3 + 1} e^{-t^2} dt \qquad &\text{Definition of backwards integrals} \tag{1} \end{align}

Next, we'll define $$\displaystyle F(x) = \int_0^x e^{-t^2} dt$$.

We know its derivative is $$F'(x) = e^{-x^2}$$, by the Fundamental Theorem of Calculus.

Next, we'll define new functions for the two integrals in $$(1)$$:

\begin{align*} H_1(x) &= \int_0^{x^2 + 2x} e^{-t^2} dt &\qquad H_2(x) &= \displaystyle\int_{0}^{x^3 + 1} e^{-t^2} dt \\ &= F(x^2 + 2x)& &=F(x^3 + 1) \end{align*}

We use the chain rule to find their derivatives:

$$H_1'(x) = e^{-(x^2 + 2x)^2} (2x + 2) \qquad H_2'(x) = e^{-(x^3 + 1)^2} (3x)$$

Therefore,

$$H'(x) = e^{-(x^2 + 2x)^2} (2x + 2) - e^{-(x^3 + 1)^2} (3x)$$

Is my calculation correct?

• A little mistake: derivative of $x^{3}+1$ is not $3x$. Jul 13, 2019 at 0:03
• The last term should be $3x^2$.
– Feng
Jul 13, 2019 at 0:03
• Correct except for the $3x$ the others mentioned. It wasn't actually necessary to split the integral in two, you could just say $F$ is any antiderivative of $\exp(-t^2)$ and write $H(x)=F(x^2+2x)-F(x^3+1)$. Jul 13, 2019 at 0:07

$$\frac{d}{dx} \int_{L(x)}^{U(x)} f(t) dt= U'(x) f(U(x)) -L'(x) f(L(x))$$ So in your case you get $$H'(x)=(2x+2) e^{-(x^2+2x)^2}-3x^2 e^{-(x^3+1)^2}.$$