Question

CALC HELP PLEASE!

Answer 1

$$
\Large {
f_{avg} = \frac{1}{b-a}\int_a^b f(x)
\\ \therefore \\
f_{avg} = \frac{1}{4-2}\int_2^4 e^{2x} ~dx

}
$$

Answer 2

ok so far?

Answer 3

yes! i was a little confused on the formula, so i just solve that?

Answer 4

yes??

Answer 5

@phi can you pleaseee continue??

Answer 6

@amistre64 please help!!

Answer 7

the rest is just working out the integral

Answer 8

what is the integration of e^(2x) ?

Answer 9

e^2x/2

Answer 10

@amistre64

Answer 11

very good, now recall the fundamental calculus thingy:

\[\int_{a}^{b}f(x)dx=F(b)-F(a)\]

Answer 12

F(x) = e^2x/2
a = 2 and b = 4

Answer 13

so it would be F(4)-F(2)

Answer 14

@amistre64

Answer 15

yes, at least the integration portion of it.

Answer 16

\[\Large {
\frac{1}{4-2}\int_2^4 e^{2x} ~dx \\
\frac{1}{2}~\frac12e^{2x}|_{2}^{4}\\
\frac{1}{4}~(e^{2(4)}-e^{2(2)})
}\]

Answer 17

so to simplify it would be 1/4 (e^8-e^4)

Answer 18

right? @amistre64 or am i completely wrong?

Answer 19

well, e^8 - e^4 might not be a pretty number, it all depends on if the question is wanting an approximation or an exactness

Answer 20

its just asking it as a decimal value

Answer 21

then i would see what the calculator gives us for e^8 - e^4 :)

Answer 22

http://www.wolframalpha.com/input/?i=%28e%5E8-e%5E4%29%2F4

Answer 23

oo that is not a pretty number, hahaha thank you!

Answer 24

just for good measure

http://www.wolframalpha.com/input/?i=integrate+e%5E%282x%29%2F2+dx%2C+from+x%3D2..4

we get the same results :)

Answer 25

thank you so much! (:

Answer 26

youre welcome, and good luck