1. ## Integral review

1) It is known that if m f(x) M for a x b, then the following property of integrals is true. $m(b-a) <= \int^b_a f(x)dx <= M(b-a)$

Use this property to estimate the value of the given integral.

____ < $\int^6_0 3xe^{-x}dx$ < ____

Would I just have to find the minimum and maximum on that interval, then plug everything in?

2)
Code:
 f(x) = {
0                   if x < 0
x                    if 0 <= x <= 1
2 - x          if 1 <= x <= 2
0                    if x > 2
}
$g(x) = \int ^x_0 f(t)dt$

(a) Find an expression for g(x) similar to the one for f(x).

State where g(x) is discontinuous (if anywhere)

I've done everything but H and K, which I think I need to decide if it's continuous or not. I thought K would just be the Integral of 0, but that's not right, and I'm not sure how to go about solving for H on this one.

2. Originally Posted by Open that Hampster!
1) It is known that if m f(x) M for a x b, then the following property of integrals is true. $m(b-a) <= \int^b_a f(x)dx <= M(b-a)$

Use this property to estimate the value of the given integral.

____ < $\int^6_0 3xe^{-x}dx$ < ____

Would I just have to find the minimum and maximum on that interval, then plug everything in?

YES

2)
Code:
 f(x) = {
0                   if x < 0
x                    if 0 <= x <= 1
2 - x          if 1 <= x <= 2
0                    if x > 2
}
$g(x) = \int ^x_0 f(t)dt$

(a) Find an expression for g(x) similar to the one for f(x).

State where g(x) is discontinuous (if anywhere)

I've done everything but H and K, which I think I need to decide if it's continuous or not. I thought K would just be the Integral of 0, but that's not right, and I'm not sure how to go about solving for H on this one.

H,K=0. btw, what is the meaning of this:
x if 0 <= x <= 1
2 - x if 1 <= x <= 2

if x=1, we can't tell whether f(x) will equal x or 2-x.

3. Originally Posted by vince
Apologies about the typo, it should be:

x if 0 <= x <= 1
2 - x if 1 < x <= 2

I forgot to mention, I've already tried 0 for both H and K. Neither were correct.

4. oh....oops.

notice that when
$g(x) = \int ^x_0 f(t)dt$ is evaluated when x is between 1 & 2, it's an intergal from 0 to 2 so that in the instance

$g(x) = \int ^x_0 f(t)dt = \int ^1_0 f(t)dt + \int ^x_1 f(t)dt$

str8forward from here right