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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.

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    $\begingroup$ Hint: chain rule. $\endgroup$ – GEdgar Nov 24 '18 at 22:34
  • $\begingroup$ As a suggested reading, check out Leibniz Rule for differentiation of integrals. $\endgroup$ – Shubham Johri Nov 24 '18 at 22:44
  • $\begingroup$ Hint: $\int_{-\infty}^x f(x) dx = f(x)$, so $\int_a^bf(x) dx = \int_{-\infty}^b f(x) dx - \int_{-infty}^a f(x) dx$ $\endgroup$ – eSurfsnake Nov 25 '18 at 7:19
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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?

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  • $\begingroup$ Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC? $\endgroup$ – James Ronald Nov 24 '18 at 22:42
  • $\begingroup$ No, the lower bound does not matter. $\endgroup$ – José Carlos Santos Nov 24 '18 at 22:45
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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 $$g^{\prime}(x) = - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}} = -\sqrt{x}\tan(\sqrt{x}) \cdot \dfrac{1}{2\sqrt{x}} = -\dfrac{1}{2}\tan(\sqrt{x})\text{.}$$

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  • $\begingroup$ Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again! $\endgroup$ – James Ronald Nov 24 '18 at 23:11

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