# ∫ 1 3 ( 2 e x − 1 ) d x

### Single Variable Calculus: Concepts...

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

### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 5.2, Problem 46E
To determine

## To evaluate : ∫13(2ex−1)dx

Expert Solution

13(2ex1)dx=2e(e21)

### Explanation of Solution

Given information : f(x)=2ex1,   where a=1,b=3

Formula used : abf(x)dx=limx(ban)i=1nf(xi)

Calculation :

We have, x= 31 n = 2 n   Here, x 0 =1

x 1 =1+ 2 n , x 2 =1+ 4 n , x 3 =1+ 6 n ,............... x i =1+ 2i n

1 3 ( 2 e x 1 )dx= lim x 2 n i=1 n ( 2 e ( 1+ 2i n ) 1 )

= lim x 2 n i=1 n ( 2 e ( 1+ 2i n ) 2 )

1 3 ( 2 e x 1 )dx= lim x 4 n i=1 n e 1+ 2i n 2     ..............................( 1 )

Now, i=1 n e 1+ 2i n =[ e 3n+2 n e n+2 n e 2 n 1 ]

Equation ( 1 ) becomes,

1 3 ( 2 e x 1 )= lim x 4 n [ e 3n+2 n e n+2 n e 2 n 1 ] = lim x 2[ e 3n+2 n e n+2 n e 2 n 1 2 n ]=2( e 3 e )=2e( e 2 1 )

Therefore, 13(2ex1)dx=2e(e21)

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!