(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]\int_{1}^{2}(\frac{2}{2x+1})^3 dx[/tex]

2. Relevant equations

Normal integral equations

3. The attempt at a solution

[tex]\int_{1}^{2}(\frac{2}{2x+1})^3 dx = 2^3\int_{1}^{2}\frac{1}{(2x+1)^3} dx[/tex]

u=2x+1

[tex]2^3\int_{1}^{2}\frac{1}{u^3} dx = 2^3\int_{1}^{2}u^{-3} dx[/tex]

Antiderivate of [tex]u^{-3} = \frac{u^{-3+1}}{-3+1} = \frac{u^{-2}}{-2}= \frac{1}{-2u^2}[/tex]

Plotting the real u in: [tex]\frac{1}{-2(2x+1)^2}[/tex]

[tex]2^3\int_{1}^{2}\frac{1}{-2(2x+1)^2} = 2^3\left(\frac{1}{-2(2x+1)^2}\right)_1^2 =2^3\left(\frac{1}{-2(2 \cdot 2+1)^2}\right) - 2^3\left(\frac{1}{-2(2 \cdot 1+1)^2}\right) = \left(\frac{2^3}{-50}\right) - \left(\frac{2^3}{-18}\right)[/tex]

[tex] = \left(\frac{72}{-450}\right) - \left(\frac{200}{-450}\right) = \left(\frac{72-200}{-450}\right) = \left(\frac{-128}{-450}\right) = \frac{128}{450} = \frac{64}{225}[/tex]

This gives about: 0.284444...

When I type this function on the calculator, the area between x=1 and x=2 is 0.1422222...

That is half of my answer, what have I done wrong?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Integration problem

**Physics Forums | Science Articles, Homework Help, Discussion**