Question

求大神解高数？

求a的值,使得y=ax*2(a大于0)与y=-x*2+4所围成的平面图形的面积为4

Answer 1

y=ax^2 (1)
y=-x^2+4 (2)
from (1) and (2)
ax^2 = -x^2 +4
(a+1)x^2 =4
x= 2/√(a+1) or -2/高缓首√(a+1)
A
=∫(-2/√(a+1) -> 2/√(a+1)) [ ( -x^2+ 4) -ax^2] dx
=2∫(0 -> 2/√戚数(a+1)) [ ( -x^2+ 4) -ax^2] dx
=2[ -(1+a)x^3/3 + 4x ]|(0 -> 2/√(a+1))
=2 { -(1+a)[2/√(a+1)] ^3/3 + 4[2/√(a+1)] }
A=4
2 { -(1+a)[2/√(a+1)] ^3/3 + 4[2/√(a+1)] } =4
-(1+a)[2/√(a+1)] ^3/3 + 4[2/√(a+1)] =2
-(2/3)[ 1/哪棚√(a+1)] + 8/√(a+1) =2
-2/3 +8 =2√(a+1)
2√(a+1) = 22/3
√(a+1) = 11/3
a+1 =121/9
a= 112/9

追问

可答案是55/9

追答

计算错误！

A=4
2 { -(1+a)[2/√(a+1)] ^3/3 + 4[2/√(a+1)] } =4
-(1+a)[2/√(a+1)] ^3/3 + 4[2/√(a+1)] =2
-(8/3)[ 1/√(a+1)] + 8/√(a+1) =2
-8/3 +8 =2√(a+1)
2√(a+1) = 16/3
√(a+1) = 8/3
a+1 =64/9
a= 55/9

追问

好的👌，谢谢