# Integration, x^3*sqrt(2-x)

• Dec 12th 2009, 02:01 PM
micro31337
Integration, x^3*sqrt(2-x)
hello, this problem has been annoying me for the past 24 hours, and i cant seem to figure out where to begin,

Evaluate the integral:
(integration sign goes here) (x^3*sqrt(2-x)) dx

ive tried substitution but got nowhere, i attempted integ. by parts, but got confused because my teacher never taught it.

this question is on my final exam review, and it says the answer is:
-(( 16/3) * (2-x)^(3/2) - (24/5) * (2-x)^(5/3) + (12/7) * (2-x)^(7/2) - (2/9)*(2-x)^(9/2)) + C

thank you.
• Dec 12th 2009, 02:07 PM
Jester
Quote:

Originally Posted by micro31337
hello, this problem has been annoying me for the past 24 hours, and i cant seem to figure out where to begin,

Evaluate the integral:
(integration sign goes here) (x^3*sqrt(2-x)) dx

ive tried substitution but got nowhere, i attempted integ. by parts, but got confused because my teacher never taught it.

this question is on my final exam review, and it says the answer is:
-(( 16/3) * (2-x)^(3/2) - (24/5) * (2-x)^(5/3) + (12/7) * (2-x)^(7/2) - (2/9)*(2-x)^(9/2)) + C

thank you.

Let $\displaystyle u = 2-x$, expand and integrate.
• Dec 12th 2009, 02:20 PM
micro31337
so u mean do substitution?
so
u = 2-x, x = 2 - u
du = -1dx
-du = dx

substituting in u: -(ingetral symbol)((2-u)^3 * u^(1/2))du
is that what you mean? i dont see how i can continue with this
• Dec 12th 2009, 02:28 PM
Jester
Quote:

Originally Posted by micro31337
so u mean do substitution?
so
u = 2-x, x = 2 - u
du = -1dx
-du = dx

substituting in u: -(ingetral symbol)((2-u)^3 * u^(1/2))du
is that what you mean? i dont see how i can continue with this

So $\displaystyle -\int (2-u)^3 \sqrt{u} du = -\int \left(8 - 12u + 6u^2 - u^3\right) u^{1/2}\,du$ $\displaystyle = \int u^{7/2} - 6 u^{5/2} + 12 u^{3/2} - 8 u^{1/2}\,du$.

Then integrate term by term.
• Dec 12th 2009, 02:41 PM
micro31337
Ooh wow, i had forgotten i could do that lol, thank you man!