Question

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

Answer 1

Use the Fundamental Theorem of Calculus Part 1 and some rewriting and the chain rule. The derivative is `sin^3(2x-3) - sin^3(2x-1)`

Explanation:

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

`g(x) = int_a^x f(t) dt`, for `x` in some interval `[a,b]` on which `f` is continuous, then we have (among other things):

`g'(x) = f(x)`

To find the derivative of

`g(x) = int_x^(x-1) f(t) dt`, where `f(t) = sin^3(2t-1)`

we note that the function is continuous on `RR`, so that is not a concern. But we need a constant as the lower limit of integration.

So we will rewrite.

`g(x) = int_x^(x-1) f(t) dt = int_x^0 f(t) dt + int_0^(x-1) f(t) dt`

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`.)

`g(x) = -int_0^x f(t) dt + int_0^(x-1) f(t) dt`

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 `x` as the upper limit of integration, so we need the chain rule version of FTC 1.

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

So for this problem we arrive at:

`g'(x) = -f(x) + f(x-1) * 1` (Of course, the derivative of `x-1` is just `1`)

So the derivative (with respect to `x`) of

`int_x^(x-1) sin^3(2t-1) dt` is:

`-sin^3(2x-1) + sin^3 (2(x-1)-1)`

The answer can be rewritten as:

`sin^3(2x-3) - sin^3(2x-1)`

Final Note

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

The derivative of `g(x) = int_(f(x))^(g(x)) h(t) dt` is

`h(g(x))*g'(x) - h(f(x))*f'(x)`.

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.