# How to approach integrals as a function?

I'm trying to solve the following question involving integrals, and can't quite get what am I supposed to do:

$$f(x) = \int_{2x}^{x^2}\root 3\of{\cos z}~dz$$ $$f'(x) =\ ?$$

How should I approach such integral functions? Am I just over-complicating a simple thing?

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Are you familiar with the Fundamental Theorem of Calculus and the chain rule for differentiation? – Tim Duff Jul 9 '12 at 20:35
Yes, I do; I tried messing with this integral around, using the Fundamental Theorem of Calculus, but wasn't quite sure I'm on the right path. I've ended up with a weird expression that I'm not sure is correct/final answer. I've tried gathering information from wolframalpha, but it doesn't seem to handle such functions/integrals. Could you direct me to the right way - what should I end with? – Dvir Azulay Jul 9 '12 at 20:40
@Sam: That looks interesting; Could you explain how it can be used here? – Dvir Azulay Jul 9 '12 at 20:41

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You could first split the integral into two integrals: $\int_{2x}^{x^2} \root3\of{\cos z}\,dz =\int_{2x}^0\root3\of{\cos z}\,dz+\int_0^{x^2}\root3\of{\cos z}\,dz=-\int_0^{2x}\root3\of{\cos z}\,dz+\int_0^{x^2}\root3\of{\cos z}\,dz$, and then use Tim's hint in his comment. – David Mitra Jul 9 '12 at 20:52

## 5 Answers

For this problem, you will ultimately use a version of the Fundamental Theorem of Calculus: If $f$ is continuous, then the function $F$ defined by $F(x)=\int_a^x f(z)\,dz$ is differentiable and $F'(x)=f(x)$.

So for instance, for $F(x)=\int_0^x\root3\of{\cos z}\,dz$, we have $F'(x)=\root3\of{\cos x}$.

One can combine this with the chain rule, when it applies, to differentiate a function whose rule is of the form $F(x)=\int_a^{g(x)} f(z)\,dz$. Here, we recognize that $F$ is a composition of the form $F=G\circ g$ with $G(x)=\int_a^x f(z)\,dz$. The derivative is $F'(x)=\bigl[ G(g(x))\bigr]'=G'(g(x))\cdot g'(x)=f(g(x))\cdot g'(x)$.

For example, for $F(x)=\int_0^{x^2}\root3\of{\cos z}\,dz$, we have $F'(x)=\root3\of{\cos x^2}\cdot(x^2)'=2x\root3\of{\cos x^2}$.

Now to tackle your problem proper and take advantage of these rules, we just "split the integral": $$\tag{1} \int_{2x}^{x^2}\root3\of{\cos z}\,dz= \int_{2x}^{0}\root3\of{\cos z}\,dz+ \int_{0}^{x^2}\root3\of{\cos z}\,dz.$$ But wait! We can only use the aforementioned differentiation rules for functions defined by an integral when it's the upper limit of integration that is the variable. The first integral in the right hand side of $(1)$ does not satisfy this. Things are easily remedied, though; write the right hand side of $(1)$ as: $$-\int_{0}^{2x}\root3\of{\cos z}\,dz+ \int_{0}^{x^2}\root3\of{\cos z}\,dz;$$ and now things are set up to use our rule (of course, you'll also use the rule $[cf+g]'=cf'+g'$).

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Such a well written answer. Wish I could up-vote it a few more times; Thank you so much – Dvir Azulay Jul 9 '12 at 23:49
@dvir, I did it for you. – Tpofofn Jul 10 '12 at 2:38

Not to give it away completely. Using the Fundamental Theorem of Calculus, $f(x) = C(x^2)-C(2x)$, where $C(x)$ is the anti-derivative of the integrand. Now, use the Chain rule to compute $f'(x)$, which will depend only on the $C'(x)$, which is the integrand itself, evaluated at $x$.

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Really thanks for your answer! – Dvir Azulay Jul 9 '12 at 23:49

\begin{eqnarray} f'(x)&=&(x^2)'(\sqrt[s]{\cos z})|_{z=x^2}-(2x)'(\sqrt[s]{\cos z})|_{z=2x}\cr &=&2x\sqrt[s]{\cos x^2}-2\sqrt[s]{\cos 2x} \end{eqnarray}

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This doesn't seem correct. – Dom Jul 9 '12 at 21:49

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Then show us what you think is correct! – Mercy Jul 9 '12 at 21:55
Shouldn't the answer be $2x\root s\of{\cos x^2} - 2\root s\of{\cos 2x}$ ? – Dom Jul 9 '12 at 22:03
Yes, you are right! – Mercy Jul 9 '12 at 22:13

Generally, to differentiate an integral of the form:

$$\int_{g_1(x)}^{g_2(x)}f(z)dz$$

we use Leibniz rule. First assume that $F(x)$ is the anti-derivative of $f(x)$. That is $F'(x) = f(x)$. Then it follows that,

$$\int_{g_1(x)}^{g_2(x)}f(z)dz = F(z)|_{z=g_2(x)} - F(z)|_{z=g_1(x)} = F(g_2(x)) - F(g_1(x))$$

Now if we differentiate this result using the chain rule we get:

$$\frac{d}{dx}\left(F(g_2(x)) - F(g_1(x))\right) = f(g_2(x))g_2'(x) - f(g_1(x))g_1'(x).$$

Note that it is not necessary to find the anti-derivative $F()$.

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Using the Leibnitz rule of differentiation of integrals, which states that if \begin{align} f(x) = \int_{a(x)}^{b(x)} g(y) \ dy, \end{align} then \begin{align} f^{\prime}(x) = g(b(x)) b^{\prime}(x) - g(a(x)) a^{\prime}(x). \end{align} Thus, for your problem $a^{\prime}(x) = 2$ and $b^{\prime}(x) = 2x$ and, therefore, \begin{align} f^{\prime}(x) = \int_{2x}^{x^2} \sqrt[3]{\cos z} dz = \sqrt[3]{\cos (x^2)} (2 x) - \sqrt[3]{\cos (2x)} (2). \end{align}

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The downvote seems a bit harsh. – user02138 Jul 16 '12 at 1:14
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