Definite Integral

**Jim Marnell** · Apr 13th 2009, 02:47 PM

$\int_{x=1}{x=0} (x^2(4x^3+2)^3dx$

$u=4x^3+2$
$du=12x^2(dx)$
$\frac{du}{12}=x^2$ ???????

Lost on what to do with these problems after this.

**Reckoner** · Apr 13th 2009, 02:57 PM

Originally Posted by Jim Marnell

$\int_{x=1}{x=0} (x^2(4x^3+2)^3dx$

$u=4x^3+2$
$du=12x^2(dx)$
$\frac{du}{12}=x^2$ ???????

Lost on what to do with these problems after this.

Let $u=4x^3+2\Rightarrow du=12x^2\,dx.$ Now we can substitute. And since this is a definite integral, we can leave things in terms of $u$ by changing the limits of integration. When $x=0,\;u=2,$ and when $x=1,\;u=6.$ So,

$\int_1^0x^2\left(4x^3+2\right)^3\,dx$

$=\int_1^0\left(4x^3+2\right)^3\left(x^2\,dx\right)$

$=\frac1{12}\int_1^0\left(4x^3+2\right)^3\left(12x^ 2\,dx\right)$

$=\frac1{12}\int_6^2u^3\,du$

$=-\frac1{12}\int_2^6u^3\,du.$

Can you finish?

**Jim Marnell** · Apr 13th 2009, 03:05 PM

Thanks! yea ill try to finish this problem up and post what i get. Just wondering why 1/12 turns into -1/12 and the 2,6 switched to 6,2 and would the du become c?

**Reckoner** · Apr 13th 2009, 03:11 PM

Originally Posted by Jim Marnell

Thanks! yea ill try to finish this problem up and post what i get. Just wondering why 1/12 turns into -1/12 and the 2,6 switched to 6,2...

$\int_a^bf(x)\,dx=-\int_b^af(x)\,dx.$

We usually evaluate definite integrals from left to right, so I switched the limits of integration. But you can apply the fundamental theorem of calculus without doing this; it works out the same way.

...and would the du become c?

I do not understand what you mean here. Are you talking about the constant of integration? The $du$ just indicates that the integral is to be evaluated with respect to $u.$

**Jim Marnell** · Apr 13th 2009, 03:15 PM

ok both explanations make sense now, thanks again!

**Jim Marnell** · Apr 13th 2009, 03:36 PM

would it become
$\frac{-1}{12}\times{\frac{u^4}{4}}+c$

then i would sub in the numbers for u then sub in the x=2 and x=6?

**Reckoner** · Apr 13th 2009, 04:13 PM

Originally Posted by Jim Marnell

would it become
$\frac{-1}{12}\times{\frac{u^4}{4}}+c$

then i would sub in the numbers for u then sub in the x=2 and x=6?

After changing the limits of integration, you do not need to back-substitute for $u.$ Just evaluate it like normal:

$-\frac1{12}\int_2^6u^3\,du$

$=-\frac1{12}\left[\frac{u^4}4\right]_2^6$

$=-\frac1{12}\left[\frac{6^4}4-\frac{2^4}4\right]$

$=-\frac1{12}(324-4)$

$=-\frac{320}{12}=-\frac{80}3$

**Jim Marnell** · Apr 13th 2009, 04:17 PM

wow for all the problems like this ive been doing, i kept resubing in the u value. Thanks for clarifying this for me. i kept wondering why i was getting these strange numbers.

**Reckoner** · Apr 13th 2009, 04:25 PM

Originally Posted by Jim Marnell

wow for all the problems like this ive been doing, i kept resubing in the u value. Thanks for clarifying this for me. i kept wondering why i was getting these strange numbers.

You can do it that way. For this problem, we would leave the limits of integration as is and get

$-\frac1{12}\int_{x=0}^{x=1}u^3\,du$

$=-\frac1{12}\left[\frac{u^4}4\right]_{x=0}^{x=1}$

$=-\frac1{12}\left[\frac{\left(4x^3+2\right)^4}4\right]_0^1$

$=-\frac1{12}\left[\frac{\left(4\cdot1^3+2\right)^4}4-\frac{\left(4\cdot0^3+2\right)^4}4\right]$

$=-\frac1{12}\left[\frac{6^4}4-\frac{2^4}4\right]$

$=-\frac{320}{12}=-\frac{80}3.$

As you can see, the answer is the same.

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