Fundamental Theorem of Calculus Part II

[Mosaness]

1. Find the derivative of:
∫cos³(t)

where a = 1/x and b = ∏/3

This was a part of a question on my first calc exam and I just wanted to know if I did it correctly.
We can solve this using the Fundamental Theorem of Calculus, Part II
The solution would be to simply plug in the values for a and b, which should give a final answer of -1 - cos³(1/x)

 

[SammyS]

1. Find the derivative of:
∫cos³(t)

where a = 1/x and b = ∏/3

This was a part of a question on my first calc exam and I just wanted to know if I did it correctly.

We can solve this using the Fundamental Theorem of Calculus, Part II

The solution would be to simply plug in the values for a and b, which should give a final answer of -1 - cos³(1/x)

What was the problem exactly ?

Was it, find $\displaystyle \frac{d}{dx}\int_{1/x}^{\pi/3}{\cos^3(t)}\,dt\ ?$

 

[Mosaness]

Yes. That was it

 

[Mosaness]

It said find the derivative of: and then the integral you gave

 

[HallsofIvy]

No, what you have is incorrect. You do NOT "just plug in values for a and b"- the problem is not quite that simple. In particular, the "Fundamental Theorem of Calculus" says that
$$\frac{d}{dx}\int_a^x f(t)dt= f(x)$$
so you have to do something about the facts that
1) "x" is in the lower limit, not the upper.
2) The limit is 1/x, not x.

$$\frac{d}{dx}\int_{1/x}^{\pi/3} cos^3(t)dt= -\frac{d}{dx}\int_{\pi/3}^{1/x}cos^3(t)dt$$
If you let u= 1/x, that is
$$-\frac{d}{dx}\int_{\pi/3}^u cos^3(t)dt= -\frac{du}{dx}\frac{d}{du}\int_{\pi/3}^u cos^3(t)dt$$

The derivative of dF/dx is (du/dx)(dF/du) so you need to find du/dx from the definition of u and dF/du from the "Fundamental Theorem of Calculus".