Asked by Jay
I'm having trouble on this question:
Find the area of the region in the first quadrant that is bounded above by the curve y= sq rt x and below by the x-axis and the line y=x -2.
Find the area of the region in the first quadrant that is bounded above by the curve y= sq rt x and below by the x-axis and the line y=x -2.
Answers
Answered by
Steve
Sketch the region. It has vertices at (0,0), (2,0), (4,2)
You can get the area by using vertical or horizontal strips. Using vertical strips, they all lie below the curve √x, but the lower boundary changes from y=0 to y=x-2 at x=2. So, the area is
∫[0,2] √x dx + ∫[2,4] √x - (x-2) dx
= 4√2/3 + 10/3 - 4/3 √2 = 10/3
Or, you could just subtract the area of the triangle from the whole area under the curve:
∫[0,4] √x dx - (1/2)(2*2)
= 16/3 - 2 = 10/3
Using horizontal strips, the left boundary is the parabola x=y^2, and the right boundary is x = y+2, so the area is
∫[0,2] (y+2)-y^2 dy = 10/3
You can get the area by using vertical or horizontal strips. Using vertical strips, they all lie below the curve √x, but the lower boundary changes from y=0 to y=x-2 at x=2. So, the area is
∫[0,2] √x dx + ∫[2,4] √x - (x-2) dx
= 4√2/3 + 10/3 - 4/3 √2 = 10/3
Or, you could just subtract the area of the triangle from the whole area under the curve:
∫[0,4] √x dx - (1/2)(2*2)
= 16/3 - 2 = 10/3
Using horizontal strips, the left boundary is the parabola x=y^2, and the right boundary is x = y+2, so the area is
∫[0,2] (y+2)-y^2 dy = 10/3
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