- #1

disregardthat

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## Homework Statement

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

## Homework Equations

Normal integral equations

## 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?

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