Question

Integration (e^x - e^-x) dx

Answer 1

And evaluate from ln (4) to 0

Answer 2

Do you know what the indefinite integral of this is going to be?

Answer 3

yes
[e^x + e^-x]

Answer 4

+C

Answer 5

Sorry

Answer 6

you got this one right. It is okay you other also do.....

Answer 7

Now I have to evaluate from ln 4 to 0. Thats where I'm having difficulty

Answer 8

you know what it is when x=0 ?

Answer 9

Plug in 0 for x.

Answer 10

e^0 is 1

Answer 11

yes

Answer 12

and e^(-0) is also 1.

Answer 13

e^0 + e^-0
= 1 + 1
=2

Answer 14

Yes

Answer 15

And now for x=ln(4)

Answer 16

then subtract 2 - (whatever you get for x=ln(4)

Answer 17

e^ln 4 + e^-ln4

Answer 18

yes ...

Answer 19

I somehow remember e^ln 4 is 4 but I want to know how

Answer 20

that will help me solve e^-ln4

Answer 21

I don't think so, ln(4) is not 4.

Answer 22

\(\large\color{black}{ e^{ln(4)}+ e^{-ln(4)}+C}\)


\(\Large\color{black}{ e^{ln(4)}+ \frac{1}{e^{ln(4)}} +C}\)


\(\Large\color{black}{ \frac{[~e^{ln(4)}~]^2}{e^{ln(4)}}+ \frac{1}{e^{ln(4)}} +C}\)


\(\Large\color{black}{ \frac{[~e^{ln(4)}~]^2+1}{e^{ln(4)}}+C}\)

Answer 23

So for x=0 you got 2+C

and for ln(4) you got \(\Large\color{black}{ \frac{[~e^{ln(4)}~]^2+1}{e^{ln(4)}}+C}\)

Answer 24

the answer given here is 9/4

Answer 25

\(\Large\color{black}{ (~\frac{[~e^{ln(4)}~]^2+1}{e^{ln(4)}}+C~)~~\color{red}{-}~~\color{blue}{(2+C)}}\)

Answer 26

The C s cancel

Answer 27

correct

Answer 28

\(\LARGE\color{black}{ \frac{[e^{ln(4)}]^2+1}{e^{ln(4) }} }\)

\(\LARGE\color{black}{ \frac{e^{2ln(4)}+1}{e^{ln(4) }} }\)

\(\LARGE\color{black}{ \frac{e^{ln(16)}+1}{e^{ln(4) }} }\)

Answer 29

this is how I would put it.

Answer 30

how did it change to 16

Answer 31

2 ln 4 = ln 4² = ln 16

Answer 32

right ?

Answer 33

ok

Answer 34

can't we solve e^ln 4 further?

Answer 35

We can approximate it, but not simplify

Answer 36

ok. thank you.

Answer 37

I forgot the 2.
It should be

\(\LARGE\color{black}{ \frac{ e^{ln(16)}+1 }{e^{ln(4)}}-2 }\)

You can re-write it as

\(\LARGE\color{black}{ \frac{ e^{ln(16)}+1-2e^{ln(4)} }{e^{ln(4)}} }\)

\(\LARGE\color{black}{ \frac{ e^{ln(16)}-2e^{ln(4)} +1}{e^{ln(4)}} }\)

\(\LARGE\color{black}{ \frac{ e^{ln(16)}-2e^{ln(4)} +1^2}{e^{ln(4)}} }\) and then using a²-b²=a²-2ab+b², you get

\(\LARGE\color{black}{ \frac{(e^{ln(4)}-1)^2}{e^{ln(4)}} }\)