How to find the derivative of a definite integral that has unusual lower and upper bounds?

Question

I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.

The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $\frac{d}{dx} \int f(x)dx = f(x)$, i.e. the first part of the fundamental theorem of calculus, to answer the question.

The function is:

$$\int_{\sqrt{x}}^{\pi/4} \theta \cdot tan\theta \cdot d\theta = g(x)$$

In words: if a function corresponds to an integral where the upper bound on the integral is $\pi/4$, the lower bound is $\sqrt{x}$, and the function being integrated is $\theta \cdot tan\theta$ with respect to $\theta$, then what is the derivative of the function?

I've tried setting $u = \pi/4$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $\sqrt{x}$, in my solution somehow, but I have no idea how.

Any help would be greatly appreciated.

Answer 1

So,$$g(x)=\int_{\sqrt x}^{\frac\pi4}\theta\tan\theta\,\mathrm d\theta=-\int_{\frac\pi4}^{\sqrt x}\theta\tan\theta\,\mathrm d\theta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?

Answer 2

After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $x$ and the lower limit is not a constant.

So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.

Suppose the lower limit is $L(x)$ and the upper limit is $U(x)$ of the integral. Define

$$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t\text{.}$$

Suppose $F$ is an antiderivative of $f$. By part II of the Fundamental Theorem of Calculus, you know that $$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t = F(U(x)) - F(L(x))\text{.}$$ Then, the derivative of $g$ is given by, assuming differentiability of $U$ and $L$,

$$\dfrac{\text{d}}{\text{d}x}[g(x)] = F^{\prime}(U(x))U^{\prime}(x)-F^{\prime}(L(x))L^{\prime}(x)$$ after making use of the chain rule for derivatives. But, $F$ is an antiderivative of $f$, so $F^{\prime} = f$, hence $$\dfrac{\text{d}}{\text{d}x}[g(x)] = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}$$ In other words, the main result is $$\boxed{ \dfrac{\text{d}}{\text{d}x}\int_{L(x)}^{U(x)}f(t)\text{ d}t = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}}$$

---

Applying to this problem, we have $f(\theta) = \theta \tan(\theta)$, $U(x) = \dfrac{\pi}{4}$, and $L(x) = \sqrt{x}$. The derivatives are $U^{\prime}(x) = 0$ and $L^{\prime}(x) = \dfrac{1}{2\sqrt{x}}$. Hence, the derivative of $g$ is $$g^{\prime}(x) = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x) = f\left(\dfrac{\pi}{4}\right)(0) - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}}$$ which simplifies to $$g^{\prime}(x) = - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}} = -\sqrt{x}\tan(\sqrt{x}) \cdot \dfrac{1}{2\sqrt{x}} = -\dfrac{1}{2}\tan(\sqrt{x})\text{.}$$