Hello!

We have to compute the integral of pi*y^2(x) dx at the given range.Note then at this range x>0 and 7-x>0.

y^(x) = 49/(7x - x^2) = 49/[x(7-x)] = 7*(1/x + 1/(7-x)).

So V = pi*int_0.75_6_[7*(1/x + 1/(7-x)]dx = 7pi*[lnx - ln(7-x)]_0.75_6 =

7pi*ln[x/(7-x)]_0.75_6 = 7pi*(ln(6) - ln(3/25)) =

7pi*ln(50).

thank you!!!!!!;))

:)

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up