Finding a derivative using the FTC Part 1

[BlackMamba]

I have a problem which asks me to find the derivative using Part 1 of the Fundamental Theorem of Calculus.

So I know that the FTC says that if:

$g(x) = \int_{a}^{h(x)} f(t) dt$ then, $g'(x) = f(h(x)) * h'(x)$

I've got what appears to be an easy problem, maybe too easy and because of that I think I'm doing something wrong. Below is the problem and my solution. If someone could just varify if I did it correctly or not, I would greatly appreciate it.

PROBLEM: $G(x) = \int_{y}^{2} sin(x^2) dx$

My Solution:
$G'(x) = sin(2^2) * 0$
$= sin(4) * 0$
$= 0$


Thanks for taking a look.

 

[AKG]

Yes, it's right. Note that G(x) basically depends only on y, so G is a constant w.r.t. x which is why it's no surprise that it's derivative w.r.t. x is 0. It's even clearer if you recognize that the x's in the integrand are just dummy variables so:

$$G(x) = \int _y ^2 \sin (x^2)\, dx$$

can be rewritten:

$$G(x) = \int _y ^2 \sin (z^2)\, dz$$

Also, the middle part of: "sin(2²)*0 = sin(4)*0 = 0" is unnecessary since regardless of the fact that 2² = 4, the whole thing is something times 0, so it's just 0.

 

[StatusX]

Are you sure it isn't G(y)?

 

[HallsofIvy]

I'm with StatusX! Any integral of "f(x)dx" with numerical limits is just a constant. The derivative of any constant function is, of course, 0.

If you are given
[tex]G(y)= \int_y^2 sin(x^2) dx[/itex]
and are asked to find
[tex]\frac{dG}{dy}[/itex]
or are given
[tex]G(x)= \int_x^2 sin(t^2)dt[/itex]
and are asked to find
[tex]\frac{dG}{dx}[/itex]
then it would be a non-trivial problem (but still easy).

 

[BlackMamba]

Thanks for all of the replies, it's greatly appreciated.