MHB Integral substitution for a polynomial function

[karush]

W.8.1.18

$$\int_{-1}^{1} \left(x^4-2x\right)^6\left(2x^3-1\right)\,dx $$

$u=x^4-2x$ $du=4x^3-2 dx =2(2 x^3-1) dx$

$$\frac{1}{2}\int_{-1}^{1} u^6\,du $$

Is this ok

 

[Guest2]

W.8.1.18

$$\int_{-1}^{1} \left(x^4-2x\right)^6\left(2x^3-1\right)\,dx $$

$u=x^4-2x$ $du=4x^3-2 dx =2(2 x^3-1) dx$

$$\frac{1}{2}\int_{-1}^{1} u^6\,du $$

Is this ok

You forgot to change the limits.

If $u = x^4-2x$ then what's $u$ when $x = -1$ and when $x = 1$?

 

[karush]

not sure what you mean, should $u$ be something else?

 

[Guest2]

not sure what you mean, should $u$ be something else?

No, what you have done is correct. However, the limits should have changed. You're integrating the first integral between $x=-1$ and $x=1$. When you make the sub $u = x^4-2x $ your limits change! So $u = x^4-2x$ and when $x = -1$ you have $u = (-1)^4-2(-1) = 1+2 = 3$ and when $x = 1$ you have $u = (1)^4-2(1) = 1-2 = -1$. So you should have got:$$\displaystyle \frac{1}{2} \int_3^{-1} u^6 \,du$$

 

[Greg]

... which is equivalent to

$$-\dfrac12\int_{-1}^3u^6\,du$$.