问题 解答题
计算下列定积分的值
(1)
π
2
0
(x+sinx)dx
 
(2)
31
(
x
+
1
x
)
2
6xdx

(3)
32
1-x
x2
dx

(4)
π
2
-
π
2
cos2xdx
答案

(1)

π
2
0
(x+sinx)dx=(
1
2
x2-cosx)|0 
π
2
=1+
π2
8

(2)

31
(
x
+
1
x
)
2
6xdx=
31
(6x2+12x+6)dx=(2x3+6x2+6x)
|31
=112;

(3)

32
1-x
x2
dx=(-
1
x
-lnx)
|32
=ln2-ln3+
1
5

(4)

π
2
-
π
2
cos2xdx=
π
2
-
π
2
cos2x+1
2
dx
=
1
2
1
2
sin2x
+x)
|
π
2
-
π
2
=
π
2

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