What is the area of the region between the graphs of y=x^2 and y = .x from x= 0 to x = 2?
a. (2/3)
b. (8/3)
c. 4
d. (14/3)
e. (16/3)
Hello,Originally Posted by frozenflames
as you may have noticed, the 2nd equation is damaged slightly. By the given borders (0 and 2) I guessed, that this equation should be linear and that the intercepting point between parabola and straight line should be (2,4). Therefore I guess that the 2nd equation should be y=2*x. If not, all following calculations are wrong!
You calculate the enclosed area as integral of the difference of functions, from one intercepting point to the next one.
The difference is here: $\displaystyle d(x)=2x-x^2$ and the intercepting points are (0,0) and (2,4).
The area is $\displaystyle \int^{2}_{0}{d(x)\ dx}=\int^{2}_{0}{\left(2x-x^2 \right) dx}=\left[x^2-\frac{1}{3}x^3 \right]^{2}_{0}=4-\frac{8}{3}=\frac{4}{3}$
This answer is missing in your list of possible answers, so my assumption about the 2nd equation must be wrong. Sorry that I couldn't help you.
Greetings
EB