Integration (1 Viewer)

Lukybear · Oct 8, 2009

$\int (\sqrt{ax}+\frac{1}{\sqrt{ax}})dx$

Very confused. Its a easy problem i know...

mpopovic09 · Oct 8, 2009

change it to ax^1/2 + 1/(ax)^1/2 or ax^-1/2 then integrate

Lukybear · Oct 8, 2009

I did that, and got
$\frac{2}{3}\sqrt{a}x^{\frac{3}{2}} + \frac{2x^\frac{1}{2}}{a^{\frac{1}{2}}}$

But thats not what the answer tells me.

Drongoski · Oct 8, 2009

Lukybear's answer already correct, except for ommision of C (not a hanging offence)

$\100dpi integral\ =\ \int \left [ \sqrt{a}\cdot x^{\frac{1}{2}}\ +\ \frac{x^{-\frac{1}{2}}}{\sqrt{a}} \right ]\ dx\ \ =\ \ \sqrt{a}\cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\ +\ \frac{1}{\sqrt{a}}\cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}}\ +\ C\\ \\ =\ \ \frac{2}{3}x\sqrt{ax}\ +\ 2\sqrt{\frac{x}{a}}\ +\ C$

Lukybear · Oct 8, 2009

thxs drongoski.

Whilst at it. Can you prove this to me?

If u and v are differentiable functions in x, prove that $\int u \frac{dv}{dx} dx = uv - \int v\frac{du}{dx} dx$

Drongoski · Oct 8, 2009

untouchablecuz · Oct 8, 2009

Integration (1 Viewer)

Lukybear

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mpopovic09

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Lukybear

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Drongoski

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Lukybear

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