Calculus Question... - Tilted Forum Project Discussion Community

krwlz · 12-09-2003, 07:53 PM

Evaluate the indefinate integral:

S (x+2)(x-7)^1/2 dx

(note. If anyones qualified to help me, you shouldnt need to be told but... anything raised to the 1/2 is the same as sq. root. Also that s represents the integrand.)

Should I be doing a "u-substitution"? Seemed like a good idea, but I cant seem to find anything to sub u in for, that makes du the rest of the problem.

Any help would be greatly appreciated.

Pragma · 12-09-2003, 09:02 PM

Are you taking the square root of everything or just the last term (x-7)? If it's just the last term, then your answer is:

Quote:

(2/5) * (x-7)^(3/2) * (x+8)

If it's the square root of everything in the integral, your answer is really big and complicated and I doubt they'd be giving you that.

http://integrals.wolfram.com/ <-- Excellent website.

Edit: To the best of my knowledge, you only use the "u substition method" (which, if memory serves, gives you Integral[ u dv ] = uv - Integral[ v du ] ) if you've got a term such as e^x or sin/cos values that do not naturally recycle themselves down to 0 under differentiation.

krwlz · 12-09-2003, 09:15 PM

ok... Yea, jus the last term was under the square root. Its hard to type that though. Thanks you tons by the way.

I just need to figure out how you got that answer... haha

rsl12 · 12-09-2003, 09:31 PM

S (x+2)(x-7)^1/2 dx

using integration by parts:

d(uv) = udv + vdu

implies that

S u*dv = uv - S v*du.

Now just figure out the variables...

u = x + 2
du = dx

v = (2/3) * (x-7)^1.5
dv = (x-7)^0.5*dx

you should be able to figure it out from that.

krwlz · 12-09-2003, 09:33 PM

That looks like its a derivative formula... Im not sure you can break up a product like that.

rsl12 · 12-09-2003, 09:36 PM

look up "integration by parts" in the index of your textbook. you should find that formula.

krwlz · 12-09-2003, 09:49 PM

Looked it up. Thanks a million man! Somehow either we never used that in class, or that was the one class I skipped. Either way, im on track now!

Thanks again.

12-09-2003, 07:53 PM	#1 (permalink)
krwlz Fledgling Dead Head Location: Clarkson U.	Calculus Question... Evaluate the indefinate integral: S (x+2)(x-7)^1/2 dx (note. If anyones qualified to help me, you shouldnt need to be told but... anything raised to the 1/2 is the same as sq. root. Also that s represents the integrand.) Should I be doing a "u-substitution"? Seemed like a good idea, but I cant seem to find anything to sub u in for, that makes du the rest of the problem. Any help would be greatly appreciated.

12-09-2003, 09:31 PM	#4 (permalink)
rsl12 On the lam Location: northern va	S (x+2)(x-7)^1/2 dx using integration by parts: d(uv) = udv + vdu implies that S udv = uv - S vdu. Now just figure out the variables... u = x + 2 du = dx v = (2/3) * (x-7)^1.5 dv = (x-7)^0.5*dx you should be able to figure it out from that. __________________ oh baby oh baby, i like gravy.

12-09-2003, 09:36 PM	#6 (permalink)
rsl12 On the lam Location: northern va	look up "integration by parts" in the index of your textbook. you should find that formula. __________________ oh baby oh baby, i like gravy.

12-09-2003, 09:15 PM	#3 (permalink)
krwlz Fledgling Dead Head Location: Clarkson U.	ok... Yea, jus the last term was under the square root. Its hard to type that though. Thanks you tons by the way. I just need to figure out how you got that answer... haha

12-09-2003, 09:33 PM	#5 (permalink)
krwlz Fledgling Dead Head Location: Clarkson U.	That looks like its a derivative formula... Im not sure you can break up a product like that.

12-09-2003, 09:49 PM	#7 (permalink)
krwlz Fledgling Dead Head Location: Clarkson U.	Looked it up. Thanks a million man! Somehow either we never used that in class, or that was the one class I skipped. Either way, im on track now! Thanks again.