How do we get to the answer?
d/dx [int (sin(t^3),dt) from t=0 to x^2]
answer: 2xsin(x^6)

Question

anonymous · Answer

Question: $$\frac{ d }{ dx } \int\limits_{0}^{x^2}\sin (t^{3})dx$$
Answer: $$2x \sin (x^{6})$$

zepdrix · Answer

$$\large \frac{d}{dx}\int\limits\limits_0^{x^2}\sin(t^3)\;dt$$

We'll have to apply the `Fundamental Theorem of Calculus, Part 1`.
Here is what the rule looks like,$$\large \frac{d}{dx}\int\limits_0^x f(t)\;dt \qquad =\qquad \frac{d}{dx}F(t)|_0^x$$Where $F(t)$ is the anti-derivative of $f(t)$.
Plugging in our limits gives us,$$\large \frac{d}{dx}\left[F(x)-F(0)\right]$$Taking the derivative gives us,$$\large f(x)-0$$That second term was just a constant, so it goes away when we differentiate.

zepdrix · Answer

Once you get comfortable with this idea, you want to skip right from the first step I took, to the last.

Because with some of these problems (Like the one they gave you here), you can actually integrate this successfully. So you need to take the integral, and then undo that integral by taking the derivative.

What changes in between those two steps though, is that the variable inside your function will change due to the limits of integration.

zepdrix · Answer

We have a small problem here though, see how your upper limit is more than just an x??
It's $x^2$.

So when we go to take that final step and take the derivative, we'll have to apply the chain rule- Multiplying by the derivative of the inner function, $x^2$.

anonymous · Answer

I understand the fundamental theorem, but my first problem came up with finding the antiderivative of sin(t^3). I believe that my current way of getting the antiderivative is rather inefficient, so (if you don't mind) could you take me through those steps?

zepdrix · Answer

It is not possible to integrate that function, that's what I'm trying to explain to you. :)

We are integrating (getting some function F(t), which we can't actually find).
We then plug in the limits of integration,
and then we take the derivative to give us back what we started with, but in terms of x instead of t now.

anonymous · Answer

Oh! That just clicked - so why would we need to further multiply the original function f(x) by the derivative of x^2?

zepdrix · Answer

It we start with an easier example (one where the anti-derivative exists), it might help to make sense of this.

anonymous · Answer

Alright (thank you in advance!)

zepdrix · Answer

$$\large \frac{d}{dx}\int\limits_0^{x^3}2t\;dt \qquad = \qquad \frac{d}{dx}t^2|_0^{x^3}$$Gives us,
$$\large \frac{d}{dx}\left[(x^3)^2-(0)^2\right]$$Taking the derivative gives us,$$\large 2(x^3)\;\frac{d}{dx}x^3$$See how we're applying the chain rule to the inner function?

zepdrix · Answer

You can see that all we really did was, swap $t$ and $x^3$, but we also had to apply the chain rule.

zepdrix · Answer

$$\large 2(\color{royalblue}{t}) \qquad \rightarrow \qquad 2(\color{royalblue}{x^3})\frac{d}{dx}x^3$$

anonymous · Answer

Yes! Ok

zepdrix · Answer

$$\large \sin(\color{royalblue}{t^3}) \qquad \rightarrow \qquad \sin(\color{royalblue}{(x^2)^3})\frac{d}{dx}x^2$$Ok so understand a little better what's happening here? :)
These types of problems show up in a few different ways, so it's important that you understand the concept.

zepdrix · Answer

$$\large \frac{d}{dx}\int\limits_x^0 f(t)dt$$This is another important one that you'll want to watch out for. See how the variable is our LOWER limit of integration? We'll end up subtracting it when we evaluate the limits, so we'll have an extra negative sign at the end.