试题与答案

求抛物线y2=x与直线x-2y-3=0所围成的平面图形的面积.

题型:解答题

题目:

求抛物线y2=x与直线x-2y-3=0所围成的平面图形的面积.

答案:

y2=x
x-2y-3=0
解得,y=-1或3.

故两个交点纵坐标分别为-1,3,

则围成的平面图形面积S=

3-1
[(2y+3)-y2]dy=(y2+3y-
1
3
y3)
|3-1
=
32
3

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