`f'(u) = (u^2 + sqrt(u))/u, f(1) = 3` Find `f`.

Textbook Question

Chapter 4, Review - Problem 70 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate f(u) using the antiderivative of the function f'(u), such that:

`int f'(u) du = f(u) + c`

`int (u^2 + sqrt u)/u du = int (u^2)/u du + int (sqrt u)/u du`

`int (u^2 + sqrt u)/u du = int u du + int u^(1/2 - 1) du `

`int (u^2 + sqrt u)/u du = u^2/2 + (u^(1/2 - 1+1))/(1/2 - 1+1) + c`

`int (u^2 + sqrt u)/u du = u^2/2 + 2sqrt u + c`

Hence, `f(u) = u^2/2 + 2sqrt u + c`

You need to evaluate the constant c, using the information f(1) = 3, such that:

`f(1) = 1^2/2 + 2sqrt 1 + c`

`3 = 1/2 + 2 + c => c = 3 - 2 - 1/2 => c = 1 - 1/2 => c = 1/2`

Hence, evaluating the function f under the given conditions yields `f(u) = u^2/2 + 2sqrt u + 1/2.`

We’ve answered 318,931 questions. We can answer yours, too.

Ask a question