Question

一道定积分数学题

已知函数f(x)=x^2,g(x)=-x^2+2(a-1)x+2a(0<=a<=1)，若曲线y=f(x)和y=g(x)所围成图形的面积为m(a).
(1)求m(a)的表达式
(2)求m(a)的取值范围

答案：（1）1/3(a+1)^3 (2)[1/3,8/3]

要详细求解过程，谢谢，尽快

Answer 1

解：
联立方程求出交点：
y=x^2
y=-x^2+2(a-1)x+2a

2x^2-2(a-1)x-2a=0
x^2-(a-1)x-a=0
(x+1)(x-a)=0
x1=-1,x2=a

根据定积分的知识可以知道：
m(a)=积分（上a,下-1)[g(x)-f(x)]dx
=积分（a,-1)(-2x^2+2(a-1)x+2a)dx
=(-2/3*x^3+(a-1)x^2+2ax|(a,-1)
=1/3*(a+1)^3

m(a)=1/3*(a+1)^3是一个单调递增的函数
m(0)=1/3,m(1)=8/3
所以：m(a)的取值范围为：[1/3,8/3]

Answer 2

答案为π／8

Answer 3

Let
x
=
a
*
siny，dx
=
a
*
cosy
dy
x
=
0，y
=
0，x
=
a，y
=
π/2
原式=
∫(0->π/2)
(a²sin²y)(acosy)
*
(acosy)
dy
=
a⁴∫(0->π/2)
(sinycosy)²
dy
=
a⁴∫(0->π/2)
(1/2
*
sin2y)²
dy
=
(a⁴/4)∫(0->π/2)
sin²2y
dy
=
(a⁴/8)∫(0->π/2)
(1-cos4y)
dy
=
(a⁴/8)
*
(y
-
cos4y)：(0->π/2)
=
(a⁴/8)
*
{[π/2
-
cos(2π)]
-
[0
-
1]}
=
(a⁴/8)
*
(π/2
-
1
+
1)
=
πa⁴/16

Answer 4

=
(a⁴/8)
*
(y
-
cos4y)：(0->π/2)...................这步错了(a⁴/8)
*
(y
-
1/4（sin4y))(0->π/2）

后面结果一样