Integration questions

[roadworx]

1. The problem statement, all variables and given/known data

I'm wondering if anyone can check these integrations for me, or suggest alternative answers if they're not quite right, or can be simplified?

1) \int x^{\sqrt{2}}

2) \int x . \sqrt{x}

3) \int \frac{1}{x^\pi}

2. Relevant equations

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3. The attempt at a solution

1) \int x^{1.4} = \frac{x^{2.4}}{2.4} + C

2) \int x . x^{\frac{1}{2}} = \frac{x^2}{2} . \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{x^{\frac{3}{2}}}{3} . x^2 + C

3) \int \frac{1}{x^\pi} = \int x^{-\pi} = \frac{x^{-\pi+1}}{-\pi+1} + C

[Vuldoraq]

The first and the third look okay to me, although in the first one I would leave the \sqrt{2} rather than putting 2.4.

In the second your almost there. When doing this sort of integral it is often easier if you simplify the integrand. By combining x*\sqrt{x} into \sqrt{x*x^2} to get {x^{3/2}} it should be easier to integrate.

[HallsofIvy]

1. The problem statement, all variables and given/known data

I'm wondering if anyone can check these integrations for me, or suggest alternative answers if they're not quite right, or can be simplified?

1) \int x^{\sqrt{2}}

2) \int x . \sqrt{x}

3) \int \frac{1}{x^\pi}

2. Relevant equations

--

3. The attempt at a solution

1) \int x^{1.4} = \frac{x^{2.4}}{2.4} + C
\sqrt{2} is NOT equal to 1.4! There is no reason to change x^{\sqrt{2}} to x^{1.4}.

2) \int x . x^{\frac{1}{2}} = \frac{x^2}{2} . \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{x^{\frac{3}{2}}}{3} . x^2 + C
An unfortunately common mistake: just as you cannot differentiate a product by just differentiating each part, you cannot integrate a product that way either. As Vuldoraq said, x\sqrt{x}= x(x^{1/2})= x^{3/2}. Integrate that.

3) \int \frac{1}{x^\pi} = \int x^{-\pi} = \frac{x^{-\pi+1}}{-\pi+1} + C
Yes, this is correct.

[roadworx]

Thanks for the replies Vuldoraq, HallsofIvy.

So, my answer should be \int x^{\frac{3}{2}} = \frac{2x^{\frac{3}{2}}}{3}

[SunGod87]

You forgot to increase the exponent by one before dividing!

[dirk_mec1]

Are you aware of the fact that you're leaving out the dx in the integrals?

[roadworx]

alright..

Second attempt.

\int x^{\frac{3}{2}} = \frac{2x^{\frac{5}{2}}}{5}

[Vuldoraq]

Thats it, much better. :smile:

To repeat what HallsofIvy said: you can't integrate products of the function being integrated in the usual manner. So you have to simplify and when you can't simplify you have to use a different method (like integration by parts or substitution, you'll come accross these later on).

May your Math prosper