complicated integral (Calculus 2)

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August 28th 2007, 05:56 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
August 28th 2007, 06:04 PM
Krizalid

Nope. Try again :):) (almost got it!!)
August 28th 2007, 06:12 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
August 28th 2007, 06:14 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
August 28th 2007, 06:15 PM
Krizalid

Correct ^^

Remember that if you're unsure about your answer, take the derivative, it should yield the integrand.
August 28th 2007, 06:16 PM
chocolatelover

Thank you very much

Regards