`f'(x) = x^(-1/3), f(1) = 1` Find `f`.

Textbook Question

Chapter 4, 4.9 - Problem 37 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate f and the problem provides f'(x), hence, you need to use the following relation, such that:

`int f'(x)dx = f(x)+ c`

You need to evaluate the indefinite integral of the power function, hence, you need to use the following formula:

`int x^(-n) dx = (x^(-n+1))/(-n+1) + c`

`int x^(-1/3) dx = (x^(-1/3+1))/(-1/3+1) + c`

`int x^(-1/3) dx =(3/2)*x^(2/3) + c`

Hence,`f(x) = (3/2)*x^(2/3) + c` . You may find c using the following information, such that:

`f(1) =1 => f(1) = (3/2)*1^(2/3) + c => 3/2 + c = 1 => c = 1 -3/2 => c = -1/2`

Hence, evaluating f(x) under the given condition, yields `f(x) = (3/2)*root(3)(x^2) - 1/2.`

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