# check evaluation of indefinate integral

• April 14th 2008, 10:43 AM
shepherdm1270
check evaluation of indefinate integral
Integral from 0 to 1: x^2 (4x^3+2)^3 dx = x^3/3(x^4)^3 = F(1) - F(0)

F(1) = 1/3
F(0) = 0

= 1/3
• April 14th 2008, 10:58 AM
earboth
Quote:

Originally Posted by shepherdm1270
Integral from 0 to 1: x^2 (4x^3+2)^3 dx = x^3/3(x^4)^3 = F(1) - F(0)

F(1) = 1/3
F(0) = 0

= 1/3

$\int_0^1\left(x^2 (4x^3+2)^3\right)dx = \frac43 \int_0^1\left(\frac34 \cdot 8x^2 (2x^3+1)^3\right) dx = \left. \frac43 \cdot \frac14 \cdot (2x^3+1)^4 \right|_0^1$

I'll leave the rest for you. (Btw the result is $\frac{80}3$ )
• April 14th 2008, 11:03 AM
Sean12345
Or you could think of it as $\int af\prime(x)~f(x)^n~=~\frac{af(x)^{n+1}}{n+1}$

Hence $\int_0^1 x^2(4x^3+2)^3 \, dx$

$= \left[ \frac{1}{48}(4x^3+2)^4 \right]_0^1$
• April 14th 2008, 11:21 AM
shepherdm1270
I'm not understanding where you got the 1/48
• April 14th 2008, 11:26 AM
shepherdm1270
Quote:

Originally Posted by earboth
$\int_0^1\left(x^2 (4x^3+2)^3\right)dx = \frac43 \int_0^1\left(\frac34 \cdot 8x^2 (2x^3+1)^3\right) dx = \left. \frac43 \cdot \frac14 \cdot (2x^3+1)^4 \right|_0^1$

I'll leave the rest for you. (Btw the result is $\frac{80}3$ )

I'm not sure how you found the anti-derivative, but I'll try to go back and figure that one out on my own...

i got F(1) = 1/3(2x^3+1)^4 = 1/3 (3)^4 = 81/3

and i got F(0) = 0

so I got 81/3..... ?
• April 14th 2008, 11:41 AM
Moo
Hello,

$F(0)=\frac{1}{3} (2*0+1)^4=\frac{1}{3}*1=\frac{1}{3}$ ;)
• April 14th 2008, 11:56 AM
Sean12345
Quote:

Originally Posted by shepherdm1270
I'm not understanding where you got the 1/48

If you think of it as $\int af\prime(x)~f(x)^n~=~\frac{af(x)^{n+1}}{n+1}$

Then $f(x)=4x^3+2 \implies f\prime(x)=12x^2$

Hence a must be $\frac{1}{12}$

Given the general result we have $\frac{\frac{1}{12}(4x^3+2)^4}{4}$

$
=~\frac{1}{48}(4x^3+2)^4
$
• April 14th 2008, 10:52 PM
earboth
Quote:

Originally Posted by shepherdm1270
I'm not sure how you found the anti-derivative, but I'll try to go back and figure that one out on my own...

...

Probably you have seen what I've done. Only in case you need some confirmation I'm going to show you how I got the anit-derivative:

$\int_0^1\left(x^2 (4x^3+2)^3\right) dx = \int_0^1\left(x^2\cdot 2^3 \cdot (2x^3+1)^3\right) dx$

Let $u = 2x^3+1~\implies~\frac{du}{dx}=6x^2~\implies~du = 6x^2 dx$

That means:

$du = \frac43 \cdot \underbrace{\frac34 \cdot 2^3 \cdot x^2}_{= 6x^2} \cdot dx$

The integral becomes:

$\int_0^1\left(\frac43 \underbrace{\cdot \underbrace{\frac34 \cdot 2^3 \cdot x^2}_{= 6x^2}}_{\frac{du}{dx}} \cdot (\underbrace{2x^3+1}_{u})^3)\right) dx = \frac43 \cdot \int_{u(0)}^{u(1)} (u^3) du$ = $\frac43 \cdot \left[\frac14 \cdot u^4\right]_{u(0)}^{u(1)}$

Now re-substitute and you'll get:
$
\int_0^1\left(x^2 (4x^3+2)^3\right) dx =\left. \frac{(2x^3+1)^4}{3}\right|_0^1$
• April 14th 2008, 11:08 PM
cigster
Introduce a new variable:
u = 4x^3 + 2

They you can get dx=du/12x^2. You insert both of these two into you initial equation and get:

x^2*u^3*du/12x^2

Due to new variable you must change limits into 2 to 6, getting final equation

u^4/48
Inserting the limits you obtain the result 26.67 or 80/3.