Your question is, evaluate:
$\displaystyle
\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
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.
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:
$\displaystyle \left| \int_{-2}^0 x dx \right| +\left| \int_{0}^2 x dx \right| =\left| \left. \frac{1}{2}x^2 \right|_{-2}^0 \right| +\left| \left. \frac{1}{2}x^2 \right|_{0}^2 \right| = 2 + 2 = 4$