Application of fundamental theorem of calculus

I have this problem:

$$\frac{d}{dx} \left( \int_{\sqrt{x}}^{x^2-3x} \tan(t) dt \right)$$

I know how found the derivative of the integral from constant $a$ to variable $x$ so:

$$\frac{d}{dx} \left( \int_a^x f(t) dt \right)$$

but I don't know how make it between two variables, in this case from $\sqrt{x}$ to $x^2-3x$

Thanks so much.

• Hint: See the Examples here about splitting the integral. Let us know if that helps. Regards Jan 21 '13 at 2:41
• Split the integral and apply the chain rule. Jan 21 '13 at 2:45
• @Amzoti thanks, thats really helped me. Jan 21 '13 at 2:58
• @calbertts: You are very welcome! I would strongly recommend going through Marvis' and André Nicolas' very nice responses too and make sure you understand them. They are both great guys and awesome contributors! Regards Jan 21 '13 at 3:02

First we work formally: you can write your integral, say $F(x)=\int_a^{g(x)}f(t)\,dt-\int_a^{h(x)}f(t)\,dt$, where $f,g$ and $h$ are the functions appearing in your problem, and $a\in\mathbb R$ is constant. Next, you can apply chain rule together with fundamental theorem of calculus in order to derivate the difference above.

What is left? the existence of such $a$: Recall that by definition the upper and lower Riemann integrals are defined for bounded functions, so it is required that your integrand $\tan$ is bounded on one of the possible two integration intervals $I=[f(x),g(x)]$ or $J=[g(x),f(x)]$. This occurs only when

$$\sqrt x,\,x^2-3x\in\Bigl(-\frac{\pi}2+k\pi,\frac{\pi}2+k\pi\Bigr),\ \style{font-family:inherit;}{\text{for some integer}}\ k\,.\tag{\mathbf{I}}$$

Since both $\sqrt{x}$ and $x^2-3x$ are continuous functions, the set of the values $x$ satisfying the previous inclusion is non-empty (easy exercise left to you) and open in $\mathbb R$, so it is a countable union of open intervals. When you try to calculate the derivative, you are working locally, that is, in some of these intervals, so you simply choose a fixed element $a$ in such interval, and proceed as stated at the beginning.

If you are not familiar with the notion of "open set", then simply solve explicitly the equation $(\mathbf{I})$ and see what happens.

Hint: Let $F(t)$ be an antiderivative of $\tan t$. We can find an explicit formula for $F(t)$, but it is better not to.

Then our integral is $F(x^2-3x)-F(\sqrt{x})$. Differentiate, using the Chain Rule, and remembering that $F'(u)=\tan u$.

In general, by chain rule, we have$$\dfrac{d}{dx} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = \dfrac{d g(x)}{dx} \dfrac{d }{d g(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right) + \dfrac{d f(x)}{dx} \dfrac{d }{d f(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right)$$ Now make use the fundamental theorem of calculus to conclude that $$\dfrac{d }{d g(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = h(g(x))$$ $$\dfrac{d }{d f(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = -h(f(x))$$ Hence, $$\dfrac{d}{dx} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = \dfrac{d g(x)}{dx} h(g(x)) - \dfrac{d f(x)}{dx} h(f(x))$$

A slightly more generalized version is the Leibniz integral rule.