The graphs intersect at (1,3/2), (100/9,5), (-4/3,5)
You can integrate over either x or y. Since one boundary is a horizontal line, integrating over x will require breaking the area into two pieces:
a = ∫[-4/3,1] 5-(6-3x)/2 dx + ∫[1,100/9] 5- 3/2 √x dx
= 49/12 + 392/27 = 2009/108
integrating over y keeps to a single region, but the widths are the difference between the two curves:
a = ∫[3/2,5] 4y^2/9 - (6-2y)/3 dy
= 2009/108
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
2y=3(x^(1/2)) , y=5 and 2y+3x=6
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