# integral

• February 20th 2008, 11:55 AM
Chez_
integral
Hi i need some help please,,
Im going integration,
i have done the first question = (ment to b a math symbol like a f with no stick)S^2 and^1 2xdx which indifferentated =2^2-1^2 = 3 and got the right answer.
But then i tried to do the others but they arent working out..
s^3 & ^0 2xdx = x^2, then 2^3-0^2 which =8 but the answe is 9..:S

But then there are some with brackets in and i dont know how to go about those help please.
s^6 ^5 (2x+1)dx ...
Thank you
• February 20th 2008, 12:29 PM
angel.white
Quote:

Originally Posted by Chez_
Hi i need some help please,,
Im going integration,
i have done the first question = (ment to b a math symbol like a f with no stick)S^2 and^1 2xdx which indifferentated =2^2-1^2 = 3 and got the right answer.
But then i tried to do the others but they arent working out..
s^3 & ^0 2xdx = x^2, then 2^3-0^2 which =8 but the answe is 9..:S

But then there are some with brackets in and i dont know how to go about those help please.
s^6 ^5 (2x+1)dx ...
Thank you

You need to work on your notation.

play around with latex in this thread.

try first writing
[PHP] $\int_{0}^{3} 2x~dx$[/PHP]
It should look like this:
$\int_{0}^{3} 2x~dx$

The \int means "integral" the _{0} means to put a zero at the bottom, the ^{3} means to put a 3 at the top. the ~ adds a space.

Anyway
$\int_0^2 2x~dx ~~~~=~~~~ [x^2]_0^3 ~~~~=~~~~(3)^2-(0)^2~~~~=~~~~3^2~~~~=~~~~9$

And for the one with brackets, just split the integral up like I did here, or you can keep them in the same group and just integrate them independently
$\int_5^6 (2x+1)~dx$

$=\int_5^6 2x~dx+\int_5^6 1~dx$

$=[x^2]_5^6~~+~~[x]_5^6$

$=[6^2-5^2]~~+~~[6-5]$

$=6^2-5^2+6-5$

$=36-25+6-5$

$=12$
• February 20th 2008, 12:29 PM
earboth
Quote:

Originally Posted by Chez_
Hi i need some help please,,
Im going integration,
i have done the first question = (ment to b a math symbol like a f with no stick)S^2 and^1 2xdx which indifferentated =2^2-1^2 = 3 and got the right answer.
But then i tried to do the others but they arent working out..
s^3 & ^0 2xdx = x^2, then 2^3-0^2 which =8 but the answe is 9..:S ..... wow!

But then there are some with brackets in and i dont know how to go about those help please.
s^6 ^5 (2x+1)dx ...
Thank you

To calculate

$\int_5^6 (2x+1)dx$ you calculate the integral of each summand:

$\int_5^6 (2x+1)dx = \left. x^2 + x \right|_5^6$

Now plug in the values and calculate the difference. You should get 12.
• February 20th 2008, 12:34 PM
angel.white
Quote:

Originally Posted by earboth
To calculate

$\int_5^6 (2x+1)dx$ you calculate the integral of each summand:

$\int_5^6 (2x+1)dx = \left. x^2 + x \right|_5^6$

Now plug in the values and calculate the difference. You should get 12.

Breathing down my neck! (Hi)
• February 20th 2008, 12:40 PM
Chez_
Hey guys thanks for all your help i have now done the 9 on the exercise but i just have these three tricky ones to do, i have tried a few things but havnt got the answer..
S^3^0 2xdx

S^5^1 xdx
and finally

S^4^-5 (x^3+3x)dx

Thank you
• February 20th 2008, 12:52 PM
angel.white
Write this one:
S^3^0 2xdx

Like this: \int_{0}^{3} 2x~dx
Highlight it, and click http://www.mathhelpforum.com/math-he...itor/latex.gif it should be located near the other options like bold, center, hyperlink, etc. If you move your mouse over a math equation, you will see what the person wrote in order to get it to work, so move your mouse over this:
$\int_{0}^{3} 2x~dx$

a little box should pop up and show you what I just said.

-----------------

$\int^{4}_{-5} (x^3+3x)~dx$

$=\int^{4}_{-5} x^3~dx~~+~~\int_{-5}^{4}3x~dx$

Can you continue from here?