# How do you calculate the derivative of #intsin^3(2t-1) dt # from #[x, x-1]#?

##### 1 Answer

#### Answer:

Use the Fundamental Theorem of Calculus Part 1 and some rewriting and the chain rule. The derivative is

#### Explanation:

The Fundamental Theorem of Calculus, Part 1 (FTC 1) tells us that for

To find the derivative of

we note that the function is continuous on

So we will rewrite.

See properties of the definite integral for

#int_a^b f(t) dt = int_a^c f(t) dt + int_c^b f(t) dt# (Pick any constant you like where I have used

#0# .)

See properties of the definite integral for

#int_a^b f(t) dt = -int_b^a f(t) dt#

We are almost there, but the second integral doesn't just have an

For

#g(x) = int_a^u f(t) dt# , we get#g'(x) = f(u) (du)/dx#

So for this problem we arrive at:

So the derivative (with respect to

The answer can be rewritten as:

**Final Note**

There are people around who use the FTC 1 in the form:

The derivative of

Memorized this way, we can skip all the set-up work I did.

I've never taught from a textbook in the US that does it this way.