Find the area between y=(3+x^2)e^2^e and thex-axis fromx =0tox =1.

I have no clue how to solve this. Please help.:)

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- Aug 22nd 2007, 04:16 PMitalygirl9area problem
Find the area between y=(3+x^2)e^2^e and the

*x-*axis from*x =*0to*x =*1.

I have no clue how to solve this. Please help.:) - Aug 22nd 2007, 04:21 PMtopsquark
- Aug 22nd 2007, 04:24 PMitalygirl9
Thats the way the problem is written. The only hint I received was

(*Hint: Use the distributive property after setting up the first integral.*)Does is make any sense to you because I am so lost. I also have another problem. Thanks for any help you can offer. :p

Find the area between f(x)=(x-3)^2 and the*x-*axis over the indicated interval, [2, 4].

- Aug 22nd 2007, 04:39 PMtopsquark
No, it doesn't make any sense to me. Well, here's how to do what you wrote:

I'm just going to set the e^2^e thing as a constant c. Whichever form it is supposed to be $\displaystyle e^{2^e}$ or $\displaystyle (e^2)^e$, it is simply a constant. So:

$\displaystyle \int_0^1(3 + x^2)c dx = c\int_0^1(3 + x^2) dx$

$\displaystyle = c \int_0^13 dx + c \int_0^1x^2 dx$

$\displaystyle = 3c \int_0^1 dx + c \int_0^1 x^2 dx$

Can you do it from here?

-Dan - Aug 22nd 2007, 04:43 PMtopsquark
$\displaystyle \int_2^4(x - 3)^2 dx$

There are two ways to do this, depending on how complicated a method you prefer.

Method 1:

$\displaystyle \int_2^4(x - 3)^2 dx = \int_2^4(x^2 - 6x + 9)dx$

$\displaystyle = \int_2^4x^2~dx - 6 \int_2^4x~dx + 9 \int_2^4 dx$

Method 2:

Let $\displaystyle y = x - 3 \implies dy = dx$

$\displaystyle \int_2^4(x - 3)^2 dx = \int_{-1}^1y^2~dy$

(Note how the limits of the integration change.)

Both of these methods, of course, give you the same answer. If you haven't done substitutions yet, ignore method 2 for now. You'll see it again in due time. :)

-Dan - Aug 22nd 2007, 04:47 PMitalygirl9
Can you explain why it was set up that way?:rolleyes: My answer I come up with is 2/3 is that correct?

- Aug 22nd 2007, 04:50 PMtopsquark
That's indeed the answer. But...

How would you set it up differently? One definition of the Riemann integral is that it is the area between the x-axis and the curve over the given domain.

Seeing as you are confused, but got the correct answer you must have done something differently. Could you tell me what you did?

-Dan - Aug 22nd 2007, 04:53 PMitalygirl9
I replaced x-3 with u to simplify it.

du=1dx

since du=1dx then dx=du/1

dx-du

used the anti-derivative

then i replaced u=x-3 and solved for x:eek:

Is this the way you did it?:p - Aug 22nd 2007, 04:54 PMtopsquark
- Aug 22nd 2007, 04:55 PMitalygirl9
What is the answer for the first problem I asked you about. I come up with 1, but i don't think thats right at all. :confused:

- Aug 22nd 2007, 04:59 PMtopsquark
- Aug 22nd 2007, 05:01 PMitalygirl9
How do you get all the symbols in your replys?:rolleyes:

- Aug 22nd 2007, 05:03 PMtopsquark
Move your mouse pointer over the symbols and left-click. You should get a box coming up with the code. Otherwise, look at the first few posts in this thread.

-Dan - Aug 22nd 2007, 05:06 PMitalygirl9
I tried with the symbols and for some reason it won't work. Can you put the solution for the 2/3 in the symbols on the computer. I have it on my scrap paper, but I would like to have the whole problem worked out on the computer? I really do thank you very much for you help with these problems. This is the last class I need to graduate with my 4 year degree. I have taken calculus twice now. :o

- Aug 22nd 2007, 05:12 PMtopsquark
If I am right, this is what you did:

$\displaystyle \int_2^4 (x - 3)^2 dx$

Let $\displaystyle u = x - 3 \implies du = dx$

Thus

$\displaystyle \int (x - 3)^2 dx = \int u^2 du$

$\displaystyle = \frac{1}{3}u^3 + C = \frac{1}{3}(x - 3)^3 + C$

So

$\displaystyle \int_2^4 (x - 3)^2 dx = \frac{1}{3}(4 - 3)^3 - \frac{1}{3}(2 - 3)^3$

$\displaystyle = \frac{1}{3} - \frac{-1}{3} = \frac{2}{3}$

-Dan