Question

Please help. Will FAN AND MEDAL!

Find the average value of f(x)=e2x over the interval [2, 4].

a. 1463.18
b. 731.59
c. 1517.78
d. 23.60

Answer 1

average value is is \[\frac{1}{b-a}\int_a^bf(x)dx\]

Answer 2

in your case \[\frac{1}{2}\int_2^4e^xdx\]

Answer 3

Okay so I just plug that into my calculator and I will give me the answer? @Satellite73

Answer 4

just integrate it
it will be \[\frac{1}{2}\left(e^4-e^2\right)\]

Answer 5

Okay so I replace e^4 with 2^4 and e^2 with 4^2 right?

Answer 6

i am sorry i told you wrong

Answer 7

you have to find the anti derivative first

Answer 8

\[\frac{1}{2}\int_2^4e^{2x}dx\]

Answer 9

i put \(e^x\) that was a mistake

Answer 10

you know the anti derivative of \(e^{2x}\)?

Answer 11

\[f(x)= \frac{ e^2x }{ 2 }+c\]

Answer 12

@Satellite73

Answer 13

@sleepyjess

Answer 14

Can you please help? @sleepyjess told me that you might be able to @perl @TheSmartOne

Answer 15

Do you have any ideas? @MrNood

Answer 16

the formula you wrote last is the correct integral

so work out the definite integral between 2 and 4

then divide by 4-2

Answer 17

Which formula? The anti-derivative one?

Answer 18

the formula you wrote last

Answer 19

substitute 4 into your equation
substitute 2 into your equation

subtract those 2 numbers
divide by 4-2

Answer 20

In the question it has [2,4]. Am I subtracting 2-4 or 4-2 and then dividing by 2-4 or 4-2? I only ask because in the problem the 2 comes before the 4. @MrNood

Answer 21

a = 2, b = 4
b-a = 4-2

Answer 22

Okay so in the equation and I replacing the e or the x with 2 and 4?

Answer 23

The x right?

Answer 24

\[ f_{avg} = \frac {1}{b-a} \int_{a}^{b} f(x) dx
\]\[ f_{avg} = \frac {1}{4-2} \int_{2}^{4} e^{2x} dx = \frac 1 2 \int_{2}^{4} e^{2x} dx
\] Now you just need to evaluate the integral and multiply by 1/2.

Answer 25

What I don't understand?

Answer 26

you haven't made any attempt to actually DO the process

substitute for X

and do what I said in my last post

Answer 27

I have attempted. I substituted for x and got this. \[f(2)=\frac{ e^2x }{ 2 }+c = e^2+c\]
\[f(4)=\frac{ e^4x }{ 2 }+c = 2e^2+c\]
\[\frac{2e^2+c-e^2+c }{ 4-2 }= \frac{ 1 }{ 2 }\]
I am not 100% sure I am correct but this is what I understood from what you said. @MrNood @perl

Answer 28

you have misread the equation

it is e^(2x) not e^2 (x)

you do not need the constant - since this is a definite integral

\[\frac{ e ^{2x} }{2 }\]

substitute x=4
substitute x= 2
subtract those 2 numbers

divide by 4-2

Answer 29

am I finding the derivative of when I substitute for x or am I just evaluating/simplifying the problem?

Answer 30

After doing that I got B. 731.59?

Answer 31

Thank you @MrNood