Question

Fundamental Theorem of Calculus

Answer 1

\[\int\limits_{0}^{x} ) \frac{ f(z)}{ (z+1)} dz = \frac{ x^2 }{ 2 }\]

Answer 2

Can someone write out how to do this so I can look at it as an example problem
Find a function f such that ...

Answer 3

No because this doesn't look like any other problems I did before.. what's the x^2/2 for?

Answer 4

I don't know how to solve this, but it's asking you to integrate f(z)/(z+1) over some range 0 to x such that the resulting expression is equal to x^2 / 2

Answer 5

start by taking the derivative of both sides

Answer 6

the integral will be some area under that expression, but it will be expressed in terms of z until you evaluate it over that specific region 0 to x

Answer 7

on the right you get \(x\) and on the left you get \(\frac{f(x)}{x+1}\)

Answer 8

I was hoping to learn how to do this... :) Help has arrived!

Answer 9

since the ft of c tells you "the derivative of the integral is the integrand"

Answer 10

now you have the equation
\[\frac{f(x)}{x+1}=x\] and you can solve for \(f(x)\) in one step

Answer 11

I don't understand what you mean by getting an x on the right

Answer 12

ok lets go slow

Answer 13

you have the equation
\[\int\limits_{0}^{x} \frac{ f(z)}{ (z+1)} dz = \frac{ x^2 }{ 2 }\] so the right hand side is \(\frac{x^2}{2}\)
when you take the derivative of the right, you get \(x\)
that is clear or no?

Answer 14

Oh okay yes but on the left it's the same thing?

Answer 15

well, they are the same function (that is what the equal sign says, right) so they must have the same derivative

Answer 16

that means the derivative of \[F(x)=\int_0^x\frac{f(z)}{z+1}dz\] is equal to the derivative of \(\frac{x^2}{2}\) i.e. the derivative of
\[\int_0^x\frac{f(z)}{z+1}dx\] is \(x\)

Answer 17

so now comes the fundamental theorem of calculus.
what is the derivative of
\[\int_0^x\frac{f(z)}{z+1}dx\]

Answer 18

x?!

Answer 19

that is one answer, yes

Answer 20

the other answer, is that by the fundamental theorem of calculus, the derivative of the integral is the "integrand" in other words that expression between the elongated S and the dz
namely
\[\frac{f(x)}{x+1}\]

Answer 21

so you just plugged the x in

Answer 22

so you know
\[\frac{f(x)}{x+1}=x\] by the fundamental theorem of calculus

Answer 23

well yes, i replaced \(z\) by \(x\) because
\[\int_0^x\frac{f(z)}{z+1}dx\] is a function of \(x\) not of \(z\)

Answer 24

it is
\[F(x)=\int_0^x\frac{f(z)}{z+1}dx\]
\[F'(x)=\frac{f(x)}{x+1}\]

Answer 25

Ah okay making connectictions. So that's all you have to do to find the function\?

Answer 26

that is what the FT of A says, if
\[F(x)=\int_a^xg(z)dz \] then
\[F'(x)=g(x)\]

Answer 27

well you didn't actually say what the question asked you just wrote
\[\int\limits_{0}^{x} \frac{ f(z)}{ (z+1)} dz = \frac{ x^2 }{ 2 }\]

but i interpreted the question to be "find \(f(x)\)"

Answer 28

Oh sorry, yes that's it. Thanks a million though you're plenty of help!

Answer 29

i get \(f(x)=x(x+1)\) from one algebra step, but the question was testing if you knew that the derivative of the integral was the integrand
yw

Answer 30

nicely explained, thanks!