# Thread: complicated integral (Calculus 2)

1. ## 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

2. Nope. Try again (almost got it!!)

3. 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

4. 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

5. Correct ^^