# a function f and a number “ a”

BuyFind

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 5.4, Problem 31E
To determine

Expert Solution

## Answer to Problem 31E

The function is f(x)=x32and the number is a=9

### Explanation of Solution

Given information: 6+axf(t)t2dt=2x

Calculation: 6+axf(t)t2dt=2x

Or, axf(t)t2dt=2x6

Now differentiating under integral sign, we get

f(x)x2.1f(a)a2.0=2.12x

Or, f(x)=x2x=x32

Therefore, 6+axf(t)t2dt=2x

Or, 6+axt32t2dt=2x

Or, 6+axt12dt=2x

Or, 6+[t12+112+1]ax=2x

Or, 6+[t1212]ax=2x

Or, 6+[2x2a]=2x

Or, 2a=6

Or, a=3

Or, a=9

Thus the function is f(x)=x32and the number is a=9

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