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Integration problem help!

Find the area between the curve y = x^2 - 3x + 3 and the line y = 3 - x. I solved these simultaneously to get the intersection as (2,1) and (0,3) but have no idea how to continue, I know I have to integrate but I'm not sure on the limits. Please help!
Original post by CleverGirl383
Find the area between the curve y = x^2 - 3x + 3 and the line y = 3 - x. I solved these simultaneously to get the intersection as (2,1) and (0,3) but have no idea how to continue, I know I have to integrate but I'm not sure on the limits. Please help!


Notice that the curve y=3−xy=3-x is greater than x2−3x+3x^2 - 3x+3 over the interval [0,2]\left[0, 2 \right]. To find the area between the curves, integrate the difference between the curves with the x-coordinates of intersection as your limits.

Area=∫02(3−x−(x2−3x+3)) dx\text{Area} = \displaystyle \int_{0}^{2} (3-x - (x^2 - 3x + 3))\ dx
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lubus
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Original post by CleverGirl383
Find the area between the curve y = x^2 - 3x + 3 and the line y = 3 - x. I solved these simultaneously to get the intersection as (2,1) and (0,3) but have no idea how to continue, I know I have to integrate but I'm not sure on the limits. Please help!


Plot your graphs if you want to be clear on the area you're integrating over, but it pretty obvious that the limits you want to use here are 2,0. (x coordinates). the formula is

F( f(x)-g(x) ) , where f(x) > g(x) in the interval [a,b]

(derive this yourself if you like)

ie f(x) is the function thats ontop.

If you cba to draw, plug in any x value into both say 1,
first equation yields 1 and the second 2. so y = 3-x > y = x^2 -3x +3 on [0,2].

Your integral would be

F( 3-x -(x^2-3x+3)) over [0,2]
(edited 10 years ago)

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