Question

How to find the integral:

Answer 1

If f is continuous and\[ \int\limits_{0}^{4}f(6x)dx+\int\limits_{0}^{12}f(2x)dx=45\], then find \[\int\limits_{0}^{24}f(x)dx\]

Answer 2

do a u substitution for each one, and change the limits accordingly. The answer will jump out like magic :)

Answer 3

The definition of integral is
\[ \int_{x_0}^{x_1} f(x) = F(x_1) -F(x_0) \]
this is what we're going to use.
\[\int_0^4 f(6x) = F(6*4) +F(6*0) = F(24)-F(0)\]
\[\int_0^{12}f(2x) = F(2*12) +F(2*0) = F(24)-F(0) \]
We put this in the original equation
\[\int_0^4 f(6x) +\int_0^{12}f(2x) = 2(F(24) -F(0)) = 45\]
and
\[ \int_0^{24} f(x) = F(24) - F(0) \]
so
\[\int_0^{24} f(x) = \frac{45}{2} \]

Answer 4

i dunno if i get something in the end that resembles 1/2*Integral [f(u)] du + 1/6*Integral [f(s)]ds = 45, can i assume it's the same as like 1/2y+1/6y = 45 and solve for y?

Answer 5

Well, f(u) and f(s) is not equal...

Answer 6

what if i set u = s?

Answer 7

u is not equal s :P

Answer 8

well like "let u = s"? as if u were doing a u-substituion kind of o_O?

Answer 9

maybe not lol. Um but if f is the same, why is f(u) and f(s0 different, aren't u just changing the letter of the variable?

Answer 10

\[f(x) = 3x\]
\[f(3x) = 3*(3x) = 9x\]
\[ f(6x) = 3*(6x) =18x \]
if you set 3x to be u and 6x to be u you get
\[ f(3x) = f(u) = 3u \]
\[ f(6x) = f(u) = 3u \]
and this is not correct.

Answer 11

Like \(f(3x) + f(6x) = 9x+18x\) but \(f(u) +f(u) = 3u+3u\)

Answer 12

im not sure how , Plitter, you got to the 2nd before last line :S in ur first post

Answer 13

well just before the last line actually (before the 45/2)

Answer 14

if you do a u substitution without an integral, that is correct, it is not the same. But with the integral there is no difference between:
\[\int\limits_{a}^{b}f(x)dx\]
or\[\int\limits_{a}^{b}f(cat)dcat\]
its a dummy variable.

Answer 15

@kirby its the definition of integral.
@joemath, i agree with what you wrote but you subsititute first u = 3x and then u = 6x. u = u but \(3x \neq 6x\)

Answer 16

@joemath what you wrote last i mean

Answer 17

my last picture shows an example of a function where:\[\int\limits_{0}^{4}f(6x)dx\neq F(24)-F(0)\]
that would only be true if the function was linear (i think so anyways, i havent tested it). I picked a quadratic function. it probably wouldnt work on any higher degree polynomial.

Answer 18

@joemath that is the definition.... http://mathworld.wolfram.com/DefiniteIntegral.html

Answer 19

@joemath if you proove this wrong I think you will have prooven the Riemann theory wrong :)

Answer 20

Can you do like set u = 2x.. and get 1/2 Integral f(u)du and set s = 6x and get 1/6 Integral f(s)ds, and then f(s)ds = f(u)du?

Answer 21

but maybe that implies that f(u) = f(s) => u = u ._. ? which wouldnt be right o.o

Answer 22

i know that is the definition, but f(6x) doesnt mean F(6x). thats wrong. It would be (1/6)F(6x), because of the chain rule.

Answer 23

u = s*

Answer 24

So your method is correct, but you forgot the 1/6 and the 1/2 that would be in front of F(6x) and F(2x) respectively.

Answer 25

oh wait actually i think the implication i stated only works if you choose particular values for f, not just substitute the variable for another variable

Answer 26

@joemath no, because I don't substitute anything.

Answer 27

Right? because:\[g(x)=6x, \frac{d}{dx}\frac{1}{6}F(g(x))=\frac{1}{6}f(g(x))\cdot g^{\prime}(x)=f(6x)\]

Answer 28

if f(x) = 2x^2 + 7
won't f(s) = 2s^2 + 7, so it's the same?

Answer 29

=( lol i am getting confused

Answer 30

@joemathIf \(g(x) = 6x\) then \(\frac{d}{dx} F(g(x)) = F'(g(x))*g'(x) = 1/6F'(g(x))\)
@kirby if you only have one function that is true, but your dealing with 2 functions here.
\[f(3x) +f(6x) = c\]
where c is a constant, then
\[f(u) + f(u) = c\] will not work.

Answer 31

\[F^{\space\prime}(x)=f(x)\]

thats straight from the Fundamental Theorem of Calc you brought up.

Answer 32

That is right, but
\[F\:'(3x) \neq F\:'(6x)\]

Answer 33

I don't see where you are going with this...

Answer 34

if you are admitting that is right, then you should see why:\[\int\limits_{0}^{4}f(6x)dx=\frac{1}{6}F(6*4)-\frac{1}{6}F(6*0)\]

Answer 35

There is no \(\frac{1}{6}\) outside. The definition of a definite integral is
\[\int_a^b f(x)dx = F(b)-F(a)\]

Answer 36

yes, but its not f(x), its f(6x).

Answer 37

it's a chain rule?

Answer 38

Heres another example where:\[\int\limits_{a}^{b}f(6x)dx\neq F(6b)-F(6a)\]
Let f(x)=sinx

Then we have:\[\int\limits_{0}^{\frac{\pi}{2}}\sin(6x)dx=-\cos(6\cdot \frac{\pi}{2})+\cos(0)=2\]

but that is wrong, the answer is one third because it should be divided by 6.
http://www.wolframalpha.com/input/?i=integral&a=*C.integral-_*Calculator.dflt-&f2=sin+6x&f=Integral.integrand_sin+6x&f3=0&f=Integral.rangestart_0&f4=pi%2F2&f=Integral.rangeend_pi%2F2&a=*FVarOpt.1-_**-.***Integral.variable---.**Integral.rangestart-.*Integral.rangeend---

Answer 39

Yeah, your right. I got the same answer as you this time. It's been interesting :)

Answer 40

it was interesting :) i really liked your method, its an interesting way to think about the problem.

Answer 41

Ok well I got confirmation from a prof and told me you can just say that f(s) = (f(u) because you have a dummy variable

Answer 42

Ok well I got confirmation from a prof and told me you can just say that f(s) = (f(u) because you have a dummy variable