I need to find the area between y=sqrt(x+3) and y = (x+3)/2
I integrated in terms of y from 0 to 2 of (y^2 -3) - (2y-3). The answer i got was -4/3, and the actual answer is 4/3. Why is it coming out this way?
Thanks for helping!
Since we're integrating with respect to y, we have to consider which curve is the 'larger' one horizontally. Sketching both curves yield that $\displaystyle y = \tfrac{1}{2}(x+3)$ is the larger one.
So: $\displaystyle A = \int_0^2 \left[2y - 3 - \left(y^2 - 3\right)\right] dx$
We can see that reversing the order is accounted for by the negative sign.
Okay thank you. I understand the concept, and I figured that was the case but I am having a hard time visualizing that 1/2(x+3) is larger. I have the functions graphed out on my paper, when I rotate the graph 90 degree counterclockwise I can see how 1/2(x+3) is larger...is this how to always look at this type of problem when integrating with respect to y?