We did an example today in class which I wanted to go through again here. The question was to calculate

$\frac{d}{dx}\int_a^{x^4}\sec t dt$

We spot the pattern immediately that it’s an FTC part 1 type question, but it’s not quite there yet. In the FTC part 1, the upper limit of the integral is just $x$, and not $x^4$. A question that we would be able to answer is:

$\frac{d}{dx}\int_a^{x}\sec t dt$

This would just be $\sec x$. Or, of course, we can show that in exactly the same way:

$\frac{d}{du}\int_a^{u}\sec t dt=\sec u$

That’s just changing the names of the variables, which is fine, right? But that’s not quite the question. So, how can we convert from $x^4$ to $u$? Well, how about a substitution? How about letting $x^4=u$ and seeing what happens. This is actually just a chain rule. It’s like if I asked you to calculate:

$\frac{d}{dx} g(x^4)$.

You would just say: Let $x^4=u$ and then we have:

$\frac{d}{dx} g(x^4)=\frac{du}{dx}\frac{d}{du}g(u)=4x^3 g'(u)$.

Doing EXACTLY the same thing in this case gives us:

$\frac{d}{dx}\int_a^{x^4}\sec t dt=\frac{du}{dx}\frac{d}{du}\int_a^{u}\sec t dt=4x^3\sec u=4x^3 \sec(x^4)$.

This now allows us to solve any such question. Have a look at the questions in this week’s tutorial and see if you can get them. Note that one other piece of information which you know, but you will need to remember is that:

$\int_a^b f(x)dx=-\int_b^a f(x) dx$.

Let me know in the comments if this post has been helpful.

Conceptually, what is really happening here is we are saying: How quickly does the area under the curve of $\sec x$ change when we change the upper limit, but the position of the upper limit is changing as $x^4$ as we vary $x$.

 How clear is this post?