# Definite Integral

• January 2nd 2009, 07:39 AM
frog09
Definite Integral
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).
• January 2nd 2009, 07:50 AM
Last_Singularity
Quote:

Originally Posted by frog09
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

You should get the answer to be 7: $\int_2^6 f(x-4) dx = 7$
• January 2nd 2009, 07:51 AM
Jester
Quote:

Originally Posted by frog09
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)
• January 2nd 2009, 07:52 AM
skeeter
Quote:

Originally Posted by frog09
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 $\int_{-2}^2 f(u) \, du = \int_{-2}^2 f(x) \, dx$

$\int_{-2}^2 f(u) \, du = \int_2^6 f(x-4) \, dx = 7$
• January 2nd 2009, 08:01 AM
Last_Singularity
LOL @ danny arrigo

Sorry
• January 2nd 2009, 08:46 AM
frog09
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?
• January 2nd 2009, 08:50 AM
Jester
Quote:

Originally Posted by frog09
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 +.