# Thread: integration with substitution HELP!

1. ## integration with substitution HELP!

hey guys, I'm trying to teach myself how to integrate with u substitution but I really don't get this:

$\displaystyle \int_{0}^{1}x^2\sqrt{x^3-1} dx$

$\displaystyle u = x^3 - 1$

$\displaystyle \frac{du}{dx} = 3x^2$

$\displaystyle du = 3x^2 dx$

$\displaystyle \frac{1}{3}\int_{0}^{1} u^\frac{1}{2} dx$

$\displaystyle = [\frac{1}{3}.\frac{2}{3}u^\frac{3}{2}]$

substituting in

$\displaystyle = \frac{2}{9}\sqrt{(x^3 - 1)^3}$

But when I substitute x = 1 and x = 0 in, it yields a negative number within the square root sign. btw, no complex numbers involved at this stage.

or have I done something wrong?

Thanks

2. Originally Posted by differentiate
hey guys, I'm trying to teach myself how to integrate with u substitution but I really don't get this:

$\displaystyle \int_{0}^{1}x^2\sqrt{x^3-1} dx$

$\displaystyle u = x^3 - 1$

$\displaystyle \frac{du}{dx} = 3x^2$

$\displaystyle du = 3x^2 dx$

$\displaystyle \frac{1}{3}\int_{0}^{1} u^\frac{1}{2} dx$

$\displaystyle = [\frac{1}{3}.\frac{2}{3}u^\frac{3}{2}]$

substituting in

$\displaystyle = \frac{2}{9}\sqrt{(x^3 - 1)^3}$

But when I substitute x = 1 and x = 0 in, it yields a negative number within the square root sign. btw, no complex numbers involved at this stage.

or have I done something wrong?

Thanks
Let $\displaystyle u = x^3 - 1$ so that $\displaystyle \frac{du}{dx} = 3x^2$.

Also note that when $\displaystyle x = 0, u = -1$, and when $\displaystyle x = 1, u = 0$

So the integral becomes

$\displaystyle \int_0^1{x^2\sqrt{x^3 - 1}\,dx} = \frac{1}{3}\int_0^1{3x^2\sqrt{x^3 - 1}\,dx}$

$\displaystyle = \frac{1}{3}\int_{x = 0}^{x = 1}{\sqrt{u}\,\frac{du}{dx}\,dx}$

$\displaystyle = \frac{1}{3}\int_{u = -1}^{u = 0}{u^{\frac{1}{2}}\,du}$

$\displaystyle = \frac{1}{3}\left[2u^{\frac{3}{2}}\right]_{-1}^0$

$\displaystyle = \frac{2}{3}\left[(-1)^{\frac{3}{2}} - 0^{\frac{3}{2}}\right]$

$\displaystyle = \frac{2}{3}\sqrt{(-1)}$...

I get the same result...

This would be because, looking at the original function $\displaystyle x^2\sqrt{x^3 - 1}$, is this function DEFINED in $\displaystyle 0 \leq x \leq 1$?

3. Thanks Prove It. umm.. I don't know. The answer at the back of the book says 1/15

4. how are you about to compute an integral where the bounds are not the correct ones? or probably the integrand is not the correct one.

note that $\displaystyle x^3-1=(x-1)(x^2+x+1),$ the second factor is clearly positive, but the second one is negative for $\displaystyle 0<x<1,$ thus we're integrating with out seein' well what we're actually integrating, hence, the integral should be $\displaystyle \int_0^1x^2\sqrt{1-x^3}\,dx$ where the integrand is well defined for the bounds.

5. Originally Posted by Krizalid
how are you about to compute an integral where the bounds are not the correct ones? or probably the integrand is not the correct one.

note that $\displaystyle x^3-1=(x-1)(x^2+x+1),$ the second factor is clearly positive, but the second one is negative for $\displaystyle 0<x<1,$ thus we're integrating with out seein' well what we're actually integrating, hence, the integral should be $\displaystyle \int_0^1x^2\sqrt{1-x^3}\,dx$ [snip]
= 2/9.

I thought that perhaps it was meant to be something like $\displaystyle \int_0^1 x^2\sqrt{1 + x^3}\,dx$ but according to the books answer of 1/15 that's not the case. Nor $\displaystyle \int_0^2 x^2\sqrt{1 + x^3}\,dx$.

So whatever the error in the question, it's nothing simple or obvious. My guess is

6. Thanks mr fantastic and krizalid for the input. so the maths book is wrong?

7. i wouldn't say it's wrong at all, i'd just say that is not the integral for the given answer, perhaps are other bounds or other integrand.