Find the area in between y^2=x-1 and y=x-3

[Darcy]

Yeah so the question is find the area between y^2=x-1 and y=x-3. I established that they intersect at (2,-1) and (5,2). The y squared threw me off so I just turned it 90 degrees anti-clockwise and got y=(x^2)+1 and y=-x+3 with x boundaries -2 and 1. Fast forward and the area is [(-(1/2)x^2)+3x](-2,1)-[((1/3)x^3)+x](-2,1) which gave 4.5 as an answer. The answers in the back of my textbook however gave 3.5. I tried it just using the original functions and integrating with respect to y and got 4.5 again. Not quite sure where I'm going wrong.

 

[pka]

Yeah so the question is find the area between y^2=x-1 and y=x-3. I established that they intersect at (2,-1) and (5,2). The y squared threw me off so I just turned it 90 degrees anti-clockwise and got y=(x^2)+1 and y=-x+3 with x boundaries -2 and 1. Fast forward and the area is [(-(1/2)x^2)+3x](-2,1)-[((1/3)x^3)+x](-2,1) which gave 4.5 as an answer. The answers in the back of my textbook however gave 3.5. I tried it just using the original functions and integrating with respect to y and got 4.5 again. Not quite sure where I'm going wrong.

Try \(\displaystyle \int_{ - 1}^2 {\left[ {\left( {y + 3} \right) - \left( {{y^2} + 1} \right)} \right]dy} \).

Look at this Your answer seems to be correct. WHY?