Solving an Improper Integral Problem: 1/(4x-2)^5 dx from 2 to infinity

loganblacke
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Homework Statement


Find the value of Integral from 2 to infinity of 1/(4x-2)^5 dx.

Homework Equations





The Attempt at a Solution


When I integrated I came up with -(1/(16(2-4x)^4)), top - bottom which ends up being 0-(-.0004822) but it says that is wrong. Help please!
 
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That looks fine to me. Maybe they want the exact answer as a fraction instead of the decimal approximation?
 
Dick said:
That looks fine to me. Maybe they want the exact answer as a fraction instead of the decimal approximation?

Good call, that is exactly what they were looking for, 1/20736. Whoever decided calculus homework should be done online should be tortured.
 
loganblacke said:
Good call, that is exactly what they were looking for, 1/20736. Whoever decided calculus homework should be done online should be tortured.

Online answer checking in general is a pain in the neck. It has no idea whether you are right or wrong, just whether it fits whatever form it's set to recognize. I sympathize.
 
loganblacke said:
When I integrated I came up with -(1/(16(2-4x)^4)), top - bottom which ends up being 0-(-.0004822) but it says that is wrong.

There should be four 0's after the decimal point: 0.000048225308641975306 ≈ 1/20736
So make sure your enter your number correctly when submitting it online. :wink:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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