So I want to try using the fundamental theorem of calculus:

$F(x) = \int_a^x f(t) dt$ defined in $[a,b]$

$F'(x) = f(x)$

I think that I have a suitable question:

Find $F'(x)$, where $F(x) = \int_0^{x^3} \sin^3t \; dt$

Now I don't understand, is $F'(x)$ actually $\frac{df}{dx}$ here because if it were $\frac{df}{dt}$ I could just take away the integral and it would be solved?

  • 1
    $\begingroup$ Note that $F(x)$ is actually $F(x^3)$ when you want to use the FTC. Now use the chain rule. $\endgroup$ – JP McCarthy Jun 25 '14 at 13:18

No, $F'(x)$ is $\dfrac{\text{d}F}{\text{d}x}$. Here's how you should approach the question.

By the first part of the fundamental theorem of calculus, you know that $$ \int\limits_{0}^{x^3}\sin^{3}(t)\text{ d}t = G(x^3)-G(0)\text{,}$$ where $G$ is an antiderivative of $\sin^{3}$; that is, $G'(x) = \sin^{3}(x)$. Take the derivative of this quantity and you get

$$\dfrac{\text{d}F}{\text{d}x} = \dfrac{\text{d}}{\text{d}x}\left[\int\limits_{0}^{x^3}\sin^{3}(t)\text{ d}t\right] = \dfrac{\text{d}}{\text{d}x}\left[G(x^3)-G(0)\right] = G'(x^3)(3x^2)-0 = G'(x^3)(3x^2)\text{.} $$ I made use of the chain rule for derivatives and that $G(0)$ is a constant and thus has derivative 0. By definition, $G'(x) = \sin^{3}(x)$ and thus your answer is $3x^2\sin^{3}(x^3)$.

A side note: This is a topic that in my four years of tutoring Calculus I material that students frequently struggle with. My experience is that once one sees how it's done once (using FTC part I and the chain rule), (s)he picks it up very quickly.

  • $\begingroup$ John claims above in his answer that $F'(x) \ne \frac{dF}{dx}$, is it? $\endgroup$ – Tony Jun 25 '14 at 21:55
  • $\begingroup$ They should be equal... as far I understand, they're equivalent notations. $\endgroup$ – Clarinetist Jun 26 '14 at 3:14

Hint :

Observe that $F(x)=H(G(x))$ where $G(x) = x^3$ and $H(x) = \int_0^x \sin^3 t \text{d}t$ and use the chain rule.


I think that you meant $\frac{dF}{dx}$ and $\frac{dF}{dt}$, respectively.

It certainly couldn't be $\frac{dF}{dt}$ because $t$ is internal to the integration operation. But, it's not $\frac{DF}{dx}$ either.

To find $F'(x)$ where $F(x) = \int_0^{x^3}\sin^3tdt$ we use the chain rule

$$\frac{d}{x^3}\int_0^{x^3}\sin^3tdt\cdot \frac{dx^3}{dx} = 3x^2\sin^3x^3$$

If you're just starting out with the the chain rule you may want to re-express it as $$\frac{d}{dx}\int_0^{u(x)}\sin^3tdt=\frac{d}{du}\int_0^{u(x)}\sin^3tdt\cdot \frac{du}{dx} = u'(x)\sin^3(u(x))=3x^2\sin^3x^3$$ where $u=x^3$

In general, suppose $$F(x)=\int_a^{u(x)}f(t)dt$$ then $$F'(x) = f(u(x))\cdot u'(x)$$

  • $\begingroup$ Should that chain rule say $sin^3 t dt \cdot \frac{dx^3}{dt}$ and not $sin^3 t dt \cdot \frac{dx^3}{dx}$? $\endgroup$ – Tony Jun 25 '14 at 23:03
  • $\begingroup$ Oh I get it, We are doing a change of variable, we want to use some $u=x^3$ and $\frac{du}{dx} = 3x^2 \to \frac{du}{3x^2} - dx$ $\endgroup$ – Tony Jun 25 '14 at 23:25
  • $\begingroup$ exactly. And don't forget that $t$ disappears after the integral has been evaluated. After evaluating the integral you are left with a function of $x$. $\endgroup$ – John Joy Jun 26 '14 at 13:16

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