Calc help please? hard problem
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Calc help please? hard problem

[From: ] [author: ] [Date: 11-05-03] [Hit: ]
.thank you so much!Using your initial condition, plugging in 0 for x and 13 for f(x),Therefore,Therefore,......
Find f(x) such that f'(x) = sqrt x -1/ sqrt x and f(0) = 13.

ALSO if you know how...

find the antiderivative
(12x^7 + 3x^5/x^4) dx

thank you so much!

-
Problem 1:

f'(x) = x^(1/2) + x^(-1/2)
Integrate f'(x)dx to get f(x) = 2/3*x^(3/2) + 2*x^(1/2) + C1
Using your initial condition, plugging in 0 for x and 13 for f(x), you get that C1 = 13

Therefore, f(x) = 2/3*x^(3/2) + 2*x^(1/2) + 13

Problem 2:
I'm going to call the expression f'(x)
f'(x) = 12x^7 + 3x NOTE x^5/x^4 = x
Integrate f'(x)dx to get f(x) = 12/8*x^8 + 3/2*x^2 + C1
Simplifying: f(x) = 3/2*x^8 + 3/2*x^2 + C1

Therefore, f(x) = 3/2*(x^8 + x^2 + C) where C = 2/3*C1
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