# Math Help - Please I need help with this integral calculus...

1. ## Please I need help with this integral calculus...

Hello...

Could you please help me out with this integral calculus... there is a though in this question...

2. Originally Posted by White Devil
Hello...

Could you please help me out with this integral calculus... there is a though in this question...
Hello,

$\int_{-2}^2 x \cdot dx = \left[\frac{1}{2} x^2 \right]_{-2}^2 = \frac{1}{2}(2^2) - \frac{1}{2}((-2)^2) = 0$

But what exactly didn't you understand?

3. Sorry, I forgot to say the answer is not 0 nor 4....

4. Originally Posted by White Devil
Hello...

Could you please help me out with this integral calculus... there is a though in this question...

$
\int_{-2}^2 x\ dx= \left. \frac{x^2}{2}\right|_{-2}^2 = 2-2 = 0
$

Now why are you having problems with this? This question is about as
simple a definite integral as you are going to see and if you are having
problems with it you need more help than just being shown the solution.

RonL

5. The answer is not 0... the teacher said there is a though in it...

6. hey white devil the answer is definitely zero and not anything else

7. Well the thing is our teacher said he will give 5 marks to who will answer it... and some students did it...

8. Hi:

I am a teacher and such status does not protect me from being wrong. Nor does it do so for your teacher (else, he/she has one terrific contract!).

Now, picture the graph of y=x on the interval [-4, 4]. As the function is symmetric about the origin, the integral from -4 to 0 is necessarily opposite that from 0 to 4. The sum of two opposites is, of course, zero.

Regards,

Rich B.

9. Originally Posted by White Devil
The answer is not 0... the teacher said there is a though in it...
The integral you gave is indeed zero, if your teacher thinks otherwise then
either they are wrong or you have not asked the question that your teacher
actually set.

Can you post what you were asked verbatim?

RonL

10. Originally Posted by White Devil
The answer is not 0... the teacher said there is a though in it...
Tell your teacher to go kill herself.

11. Originally Posted by White Devil
The answer is not 0... the teacher said there is a though in it...
Hello,

it's me again. If the answer isn't zero the only possibility I can think of is that you should calculate the area which is enclosed by y = x and x = -2 and x = 2.

The straight line with the equation y = x has a zero at x = 0. Thus you have to split the integral into 2 parts:

$\left| \int_{-2}^0 x dx \right| +\left| \int_{0}^2 x dx \right| =\left| \left. \frac{1}{2}x^2 \right|_{-2}^0 \right| +\left| \left. \frac{1}{2}x^2 \right|_{0}^2 \right| = 2 + 2 = 4$

12. earboth you're right I have asked in other forum (Arabic one) and they told me the answer and it's exactly as you did...

But could you please make it graphical?

13. Originally Posted by ThePerfectHacker
Tell your teacher to go kill herself.
Please consider youself to have been given an infraction for inappropriate
language

RonL

14. Originally Posted by White Devil
earboth you're right I have asked in other forum (Arabic one) and they told me the answer and it's exactly as you did...

But could you please make it graphical?
Hello,

I'm not quite certain what you are asking for. I've attached a diagram showing the area in question:

15. Thank you earboth that's what I asked for... Thanks all...