Definite Integral

**frog09** · January 2nd 2009, 07:39 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** · January 2nd 2009, 07:50 AM

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 $a<b<c \in \mathbb{R}$

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

**Danny** · January 2nd 2009, 07:51 AM

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)

**skeeter** · January 2nd 2009, 07:52 AM

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

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

**Last_Singularity** · January 2nd 2009, 08:01 AM

LOL @ danny arrigo

Sorry

**frog09** · January 2nd 2009, 08:46 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** · January 2nd 2009, 08:50 AM

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 +.

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