# Thread: [SOLVED] Definite Intgral Involving lnx

1. ## [SOLVED] Definite Intgral Involving lnx

This is the equation:

$\frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt$

and this is what I have:

$(4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}$

Now where do I go from here?

2. Originally Posted by myngo9191
This is the equation:

$\frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt$

and this is what I have:

$(4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}$

Now where do I go from here?
This follows from the fundemental theorem of calculus and what have there is not correct.

$\frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g(x)$

Can you compute knowing the above?

3. Originally Posted by AllanCuz
This follows from the fundemental theorem of calculus and what have there is not correct.

$\frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g(x)$

Can you compute knowing the above?
Is this what I should have?

$(4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2}$

4. Originally Posted by myngo9191
Is this what I should have?

$(4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2}$
From above,

$\frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x}$

We do not evaluate here which is what you have in your post. We are finished, this is the answer!

5. Originally Posted by AllanCuz
From above,

$\frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x}$

We do not evaluate here which is what you have in your post. We are finished, this is the answer!

0o0h!!! Thank you s0o0 much!!!