Fundamental theorem to evaluate derivative

Let's say that I were to walk up to you on the street and said, "All right, I have this function g of x which I'm going to define as the definite integral from 19 to x of the cube root of t dt."

And then I were to ask you, "What is the derivative of g evaluated at 27?"

Pause this video and think about how you would approach that.

Well, the most obvious thing would be, "All right, g prime of x." And I'll switch colors here just for a little bit of contrast.

g prime of x is going to be the derivative with respect to x of all of this business, of all of—I’ll put it in brackets here—the integral from 19 to x of the cube root of t dt.

And you might be tempted to take the anti-derivative here and evaluate it at x, evaluate it at 19, find the difference, and then take the derivative.

And you could do that and then evaluate that at 27. But we're going to see a shortcut.

You could just use the fundamental theorem of calculus.

If you're taking the derivative with respect to x of a function that's defined by a definite integral, right, like this, where our upper boundary is x, this is just going to be equal to our inner function right over here, with instead of t being the variable, it would now be x.

So this is just going to be equal to the cube root of x, and that saved us some time there.

But it can be really useful if this function inside the integral is really hard to take the anti-derivative of.

It's really hard to evaluate this integral. The fundamental theorem of calculus can be very useful.

And so going back to our original question, g prime of 27, well, this is just going to be equal to the cube root of 27, which of course is equal to 3.

And we're done.