问题
解答题
计算: (1)求函数y=
(2)
|
答案
(1)∵y=
-x
sinx+e-x1 2 ∴y′=
-1 2 x
cosx-e-x1 2
(2)原式=
(1-x)dx+∫ 10
(x-1)dx=(x-∫ 21
)x2 2
+(| 10
-x)x2 2 | 21
=(
-0)+[0-(-1 2
)1 2
=1.
计算: (1)求函数y=
(2)
|
(1)∵y=
-x
sinx+e-x1 2 ∴y′=
-1 2 x
cosx-e-x1 2
(2)原式=
(1-x)dx+∫ 10
(x-1)dx=(x-∫ 21
)x2 2
+(| 10
-x)x2 2 | 21
=(
-0)+[0-(-1 2
)1 2
=1.