Question

need with following problem

Answer 1

You might want to type your problem and post it here.

Answer 2

I need help finding the unique anti-derivative F of the top equation for which F(0)=0

Answer 3

we can do this

Answer 4

thanks! This problem is to hard for me

Answer 5

even though it is an annoying word document.
first of all divide, and get
\[2e^x+4+4e^{-3x}\]

Answer 6

let me get to a better computer

Answer 7

ok

Answer 8

ok there. was the first step clear? i just divided everything by
\[e^{2x}\]

Answer 9

yes

Answer 10

ok now we need to find the antiderivative of all this mess which we do term by term. fortunately it is not too hard in this case

Answer 11

the antideivative of
\[2e^x\] is
\[2e^x\]
the antiderivative of 4 is
\[4x\] and the antiderivative of
\[4e^{-x}\] is
\[-4e^{-x}\] so far so good?

Answer 12

why is the last one negative?

Answer 13

giving a total antiderivative of
\[F(x)=2e^x+4x-4e^{-x}+C\] good question

Answer 14

first of i assume you mean why is the antiederivative negative.

Answer 15

correct that is my question

Answer 16

the derivative of
\[e^x\] is
\[e^x\] but the derivative of
\[e^{-x}\] is
\[-e^{-x}\] by the chain rule

Answer 17

so if i want a function whose derivative is
\[e^{-x}\] i have to put a minus sign in front other wise it will not work

Answer 18

the check is easy. the derivative of
\[-4e^{-x}\] is
\[4e^{-x}\] right? and that is what we want

Answer 19

but how come the 2e^x is not negative?

Answer 20

ok hold on i think i made a mistake. we got as far as
\[f(x)=2e^x+4+4e^{-3x}\] right?

Answer 21

yes

Answer 22

and we want
\[F(x)\]

Answer 23

yes

Answer 24

ok so what function has derivative
\[2e^x\]?

Answer 25

the function that has derivative
\[2e^x\] is just
\[2e^x\] it is its own derivative

Answer 26

so you didn't have to change nothing?

Answer 27

not for that one because it is its own derivative

Answer 28

the antiderivative of
\[4\] is
\[4x\] right? because the derivative of
\[4x\] is \[4\]

Answer 29

ok I get it

Answer 30

now the only hard one is
\[4e^{-3x}\]

Answer 31

so were still not done because we need to find f(0)=0

Answer 32

oh we are not done because we also still didn't find the anti- derivative of
\[4e^{-3x}\] . this is a u-substitution or the chain rule backwards. if i take the deriviative of
\[e^{-3x}\] i get
\[-3e^{-3x}\] by the chain rule. so in order to compensate when finding the anti derivative we have to DIVIDE BY -3

Answer 33

so the anti derivative of
\[4e^{3-x}\] is
\[-\frac{3}{4}e^{-3x}\] and the check is to take the derivative and see that it works

Answer 34

now we get
\[F(x)=2e^x+4x-\frac{4}{3}e^{-3x}+C\] and we want to make sure that
\[F(0)=0\]

Answer 35

so here is the idea: replace x by 0 and solve for C

Answer 36

\[
f(0)=2e^0+4\times 0-\frac{4}{3}e^0+C=2-\frac{4}{3}+C=\frac{2}{3}+C\]

Answer 37

and if this is going to equal 0 put
\[0=\frac{2}{3}+C\] so
\[C=-\frac{2}{3}\]

Answer 38

I still confuse on the u-substitution part

Answer 39

for a "final answer" of
\[F(x)=2e^x+4x-\frac{4}{3}e^{-3x}-\frac{2}{3}\]

Answer 40

ok lets do another example

Answer 41

ok

Answer 42

what is the anti derivative of cosine?

Answer 43

positive sine

Answer 44

right.

Answer 45

now, suppose instead of wanting the antiderivative of
\[\cos(x)\] i want the anti derivative of
\[\cos(4x)\]

Answer 46

i can't say
\[\sin(4x)\] because that will not work

Answer 47

the derivative of
\[\sin(4x)\] is
\[4\sin(4x)\]

Answer 48

but we can adjust by dividing by 4. on other words i say the anti derivative of
\[\cos(4x)\] is
\[ \frac{1}{4}\sin(4x)\] and now it works

Answer 49

so 4sin(4x), is not the correct format?

Answer 50

usually this is written as
\[\int\cos(4x)dx\] i say put
\[u=4x\]
\[du=4dx\]
\[\frac{1}{4}du=dx\] and then write
\[\int\frac{1}{4}\cos(u)du=\frac{1}{4}\sin(u)=\frac{1}{4}\sin(4x)\]

Answer 51

no when you are going backwards from derivative to anti derivative you have do divide to make it work

Answer 52

OK! If you could show me the u-substitution for the problem we were working with I would appreciate it

Answer 53

clear right because the derivative of
\[\frac{1}{4}\sin(4x)\] is
\[\frac{1}{4}\cos(4x)\times 4=\cos(4x)\]

Answer 54

sure post i will try to answer

Answer 55

anti derivative to 4e^-3x

Answer 56

ok this is the one we just did. you want
\[\int4e^{-3x}dx=4\int e^{-3x}dx\]put
\[u=-3x\]
\[du=-3dx\]
\[-\frac{1}{3}du=dx\] and rewrite as
\[-\frac{4}{3}\int e^udu\]
\[=-\frac{4}{3}e^u=-\frac{4}{3}e^{-3x}\]

Answer 57

I think I finally getting the hang of this

Answer 58

good. we can do another if you like