6
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ A little mistake: derivative of $x^{3}+1$ is not $3x$. $\endgroup$ Jul 13, 2019 at 0:03
  • 1
    $\begingroup$ The last term should be $3x^2$. $\endgroup$
    – Feng
    Jul 13, 2019 at 0:03
  • 1
    $\begingroup$ 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 $H(x)=F(x^2+2x)-F(x^3+1)$. $\endgroup$
    – runway44
    Jul 13, 2019 at 0:07

1 Answer 1

3
$\begingroup$

Use Lebnitz rule:

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

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.