   Chapter 4.9, Problem 6E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the most general antiderivative of the function. (Check your answer by differentiation.)f(x) = (x − 5)2

To determine

To find: The most general antiderivative of the function f(x)=(x5)2 and check the determined antiderivative for the function f(x)=(x5)2 by differentiation.

Explanation

Given Data:

Write the given function as follows.

f(x)=(x5)2

Formula used 1:

The antiderivative function for the function xn is xn+1n+1+C .

Here, C is the constant.

Formula used 2:

Write the required differentiation formula to verify the answer as follows.

ddx(xn)=nxn1ddx(constant)=0

Calculation:

Rewrite the function f(x)=(x5)2 as follows.

f(x)=x2+5210x

f(x)=x210x1+25x0 (1)

From the antiderivative function formula, the antiderivative function for the function in equation (1) is written as follows.

F(x)=x2+12+110(x1+11+1)+25(x0+10+1)+C=x3310(x22)+25x+C=13x35x2+25x+C

Thus, the most general antiderivative of the function f(x)=(x5)2 is 13x35x2+25x+C_

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