# complicated integral (Calculus 2)

• Aug 28th 2007, 06:56 PM
chocolatelover
complicated integral (Calculus 2)
Hi everyone,

Can you please tell me if this is correct and if it isn't can you show me what I did wrong?

Find the interval:

y^3 square root of (2y^4-1)dy

u=2y^4-1
du=8y^3dy
y^3dy=du/8

y^3square root of 2y^4-1dy=
square root of u du/8=
square root of u/8=
u^(3/2)/9+c=
1/9(2y^4-1)^(3/2)+c

Thank you very much
• Aug 28th 2007, 07:04 PM
Krizalid
Nope. Try again :):) (almost got it!!)
• Aug 28th 2007, 07:12 PM
Jhevon
Quote:

Originally Posted by chocolatelover
Hi everyone,

Can you please tell me if this is correct and if it isn't can you show me what I did wrong?

Find the integral:

y^3 square root of (2y^4-1)dy

u=2y^4-1
du=8y^3dy
y^3dy=du/8

y^3square root of 2y^4-1dy=
square root of u du/8=
square root of u/8=
u^(3/2)/9+c= ...............................Here's your mistake. this integral is not correct
1/9(2y^4-1)^(3/2)+c

Thank you very much

$\int y^3 \sqrt { 2y^4 - 1}~dy$

We proceed by substitution

Let $u = 2y^4 - 1$

$\Rightarrow du = 8y^3~dy$

$\Rightarrow \frac {1}{8}du = y^3~dy$

So our integral becomes

$\frac {1}{8} \int \sqrt {u}~du$

$= \frac {1}{8} \int u^{1/2}~du$

Now continue. Hint: use the power rule
• Aug 28th 2007, 07:14 PM
chocolatelover
Hi,

Is it 1/12(2y^4-1)^3/2+c?

This is what I did:

u=2y^4-1
du=8y^3dy
y^3dy=du/8

y^3 square root (2y^4-1)dy=

square root of u du/e=
1/8 square root of u du=
1/8u^3/2/3/2=
u^3/2/12+c=
1/12(2y^4-1)^3/2+c

Thank you
• Aug 28th 2007, 07:15 PM
Krizalid
Correct ^^