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

1 Answer
Jul 6, 2015

Answer:

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.