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?
