What is the derivative of the integral of x^2t^2 from 0 to x?

Appa
Messages
15
Reaction score
0

Homework Statement



d/dx (\int^{x}_{0} x2t2dt)
So the problem is to solve the derivative of the integral \int x2t2dt from 0 to x.

Homework Equations



d/dx (\int^{x}_{a} f(t)dt) = f(x)

The Attempt at a Solution



I'm really unsure of how this should be computed but this was my guess:

d/dx (\int^{x}_{0} x2t2dt) = d/dx (1/3x2(x)3 -(1/3x2(0)3)) = d/dx (1/3x5) = 5/3x4

So, first I calculated the integral with respect to t and then derivated it with respect to x. But it feels wrong. I don't know how to treat the function x2t2 because the variable x is both part of the function and an endpoint of the interval for integration.
 
Physics news on Phys.org
You don't have to treat x^2 as 'part of the function'. It's just a constant multiplying the function t^2. You can take it out of the integral. If you have to do more complicated problems where the x and t are mixed up so you can't do that, check out the Leibniz Integral Rule. So your answer is correct.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top