Definite Integral

frog09 · #1 $Old$ January 2nd, 2009, 09:39 AM

If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).

Last_Singularity · #2 $Old$ January 2nd, 2009, 09:50 AM

Quote:

Originally Posted by frog09 $View Post$

If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).

Here is a hint: what does substituting $4-x$ instead of $x$ do to the function? An easy example to find out is to graph $f(x) = x^2$ and then graph $f(x-4) = (x-4)^2$ . Compare $\int_2^6 x^2 dx$ and $\int_6^{10} (x-4)^2 dx$

Do you see why? What I used a change of variables for the second integral, where $y=x-4,y+4=x$ and $dy = dx$ ?

Once you figure that out, recall that for that integral is a linear operator:
$\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx$ where $a<b<c \in \mathbb{R}$

You should get the answer to be 7: $\int_2^6 f(x-4) dx = 7$

Danny · #3 $Old$ January 2nd, 2009, 09:51 AM

Quote:

Originally Posted by frog09 $View Post$

If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).

Hint: Let $t = 4-x$ in $\int^6_2 f(4-x)\;dx.$ (Too slow)

skeeter · #4 $Old$ January 2nd, 2009, 09:52 AM

Quote:

Originally Posted by frog09 $View Post$

If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).

$\int_{-2}^6 f(x) \, dx = 10$

$\int_2^6 f(x) \, dx = 3$

$\int_2^6 f(x-4) \, dx = \, ?$

use the method of substitution ...

$u = x - 4$

$du = dx$

substitute and reset the limits of integration ...

$\int_2^6 f(x-4) \, dx$

$\int_{-2}^2 f(u) \, du$

$\int_{-2}^6 f(x) \, dx - \int_2^6 f(x) \, dx = \int_{-2}^2 f(x) \, dx$

$10 - 3 = 7$

since $\int_{-2}^2 f(u) \, du = \int_{-2}^2 f(x) \, dx$

$\int_{-2}^2 f(u) \, du = \int_2^6 f(x-4) \, dx = 7$

Last_Singularity · #5 $Old$ January 2nd, 2009, 10:01 AM

LOL @ danny arrigo

Sorry

frog09 · #6 $Old$ January 2nd, 2009, 10:46 AM

I was re-checking this problem and realized I copied down the last part wrong. It is really supposed to be:
$\int_2^6 f(4-x) , dx = ?$

i substitute u=4-x and get -du=dx. and get the new limits as
$\int_2^{-2} f(u) du$

which would be
$-\int_{-2}^2 f(u) du$

does that -du change the problem? or would it still be 7?

Danny · #7 $Old$ January 2nd, 2009, 10:50 AM

Quote:

Originally Posted by frog09 $View Post$

I was re-checking this problem and realized I copied down the last part wrong. It is really supposed to be:
$\int_2^6 f(4-x) , dx = ?$

i substitute u=4-x and get -du=dx. and get the new limits as
$\int_2^{-2} f(u) du$

which would be
$-\int_{-2}^2 f(u) du$

does that -du change the problem? or would it still be 7?

You would get a negative from the dx so your second last line should be - and the last line should be +.