Definite Integral (x/[sqrt(x-6)])dx

MarkSA

Junior Member
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Sep 8, 2007
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243
Hello,

I'm doing a practice worksheet, and i'm getting incorrect answers for some integrals. I was wondering if someone can maybe look at this and identify where i'm going wrong at?

15) Find the integral from 7 to 10 of: (x/[sqrt(x-6)])dx
Let u = x-6, du = dx, x = u + 6
Integral from 7 to 10 of: (u + 6)*u^(-1/2)du
Integral from 7 to 10 of: [u^(1/2) + 6u^(-1/2)]du
= 2/3*u^(3/2) + 12u^(1/2) (No +C is needed here since it's a definite integral, right?)
= 2/3*(x - 6)^(3/2) + 12(x-6)^(1/2)
F(10)-F(7): 2/3*(4)^(3/2) + 12(1)^(1/2)
= 16/3 + 12 = 52/3

The answer is supposed to be 50/3... Can anyone see any mistakes with my work? Or is the answer sheet wrong on this one?

By the way, is it possible to check your work for definite integrals in some way (aside from doing the entire problem over?) like you can with indefinite integrals by differentiating the answer?
 
MarkSA said:
Hello,

I'm doing a practice worksheet, and i'm getting incorrect answers for some integrals. I was wondering if someone can maybe look at this and identify where i'm going wrong at?

15) Find the integral from 7 to 10 of: (x/[sqrt(x-6)])dx
Let u = x-6, du = dx, x = u + 6
Integral from 7 to 10 of: (u + 6)*u^(-1/2)du
Integral from 7 to 10 of: [u^(1/2) + 6u^(-1/2)]du
= 2/3*u^(3/2) + 12u^(1/2) (No +C is needed here since it's a definite integral, right?)
= 2/3*(x - 6)^(3/2) + 12(x-6)^(1/2)
F(10)-F(7):

[(2/3)*4^(3/2) + 12*4^(1/2)] - [(2/3)*1^(3/2) + 12*1^(1/2)] =
[16/3 + 24] - [2/3 + 12] =
14/3 + 12 = 50/3


The answer is supposed to be 50/3... Can anyone see any mistakes with my work? Or is the answer sheet wrong on this one?

By the way, is it possible to check your work for definite integrals in some way (aside from doing the entire problem over?) like you can with indefinite integrals by differentiating the answer?
 
Wow, not sure how I messed up on that!

Probably is a bad idea to do integrals in the morning before coffee :)

By the way, is it possible to check your work for definite integrals in some way (aside from doing the entire problem over?) like you can with indefinite integrals by differentiating the answer? I'm guessing no but thought i'd ask
 
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