内容正文:
数学 (选修 2 - 2·人教 A 版)
| x - 3 | = x - 3,x ≥ 3,
3 - x,x < 3,{
所以∫20 ( | x - 1 | +| x - 3 | )dx
= ∫20 | x - 1 | dx + ∫
2
0
| x - 3 | dx = ∫10 (1 - x)dx + ∫
2
1
(x - 1)dx +
∫20 | x - 3 | dx
= ∫10 (1 - x)dx + ∫
2
1
(x - 1)dx + ∫20 (3 - x)dx
= (x - 1
2
x2 ) | 10 + (
1
2
x2 - x) | 21 + (3x -
1
2
x2 ) | 20
= 1
2
+ 1
2
+ 4 = 5.
跟踪练习 3:(1)C ∫20 f(x)dx = ∫
1
0
x2 dx + ∫21 (2 - x)dx
= 1
3
x3 | 10 + (2x -
1
2
x2 ) | 21 =
1
3
+ 1
2
= 5
6
.
(2)3 ∫
3π
2
0
| sinx | dx = ∫π0 sinxdx + ∫
3π
2
π
( - sinx)dx
= ( - cosx) | π0 + cosx
3π
2
π
= 2 + 1 = 3.
典例试做 4:∵ [(3t2 - 2t + 1)x]′ = 3t2 - 2t + 1,
∴ ∫31 (3t2 - 2t + 1)dx = [(3t2 - 2t + 1)x] | 31 = 3(3t2 - 2t + 1) - (3t2
- 2t + 1) = 6t2 - 4t + 2.
课堂达标·固基础
1. B ∫21 (x + 1x )dx = (
1
2
x2 + lnx) | 21 = 2 + ln2 -
1
2
= ln2 + 3
2
;
故选 B.
2. 2 ∫
π
2
- π2
(sinx + cosx)dx = ( - cosx + sinx) |
π
2
- π2
= 1 + 1 = 2.
3. 1 ∫21 (2ax + a + 1)dx = (ax2 + ax + x) | 21 = 4a + 1,所以 4a + 1 = 5,
所以 a = 1.
4. (1)∵ | x - 1 | +| x - 2 | =
3 - 2x,0 ≤ x ≤ 1,
1,1 < x ≤ 2,
2x - 3,2 < x ≤ 3,
{
∴ ∫30 ( | x - 1 | +| x - 2 | )dx = ∫
1
0
(3 - 2x)dx + ∫21 1dx + ∫
3
2
(2x - 3)dx
= (3x - x2 ) | 10 + x |
2
1 + (x
2 - 3x) | 32 = 5.
(2)∵ [ 1
2
sin(2x + π
6
)]′
= 1
2
cos(2x + π
6
)·(2x + π
6
)′ = cos(2x + π
6
),
∴ ∫
π
2
0
cos(2x + π
6
)dx = 1
2
sin(2x + π
6
) |
π
2
0
= 1
2
[sin(2 × π
2
+ π
6
) - sin(2 × 0 + π
6
)]
= 1
2
(sin 7π
6
- sin π
6
) = - 1
2
.
1. 7 定积分的简单应用
新知导学
1. (1) ∫ba f(x)dx - ∫
b
a
f(x)dx - ∫ca f(x)dx + ∫
b
c
f(x)dx
(2)∫ba [f(x) - g(x)]dx ∫
b
a
[f(x) - g(x)]dx
2. ∫ba v(t)dt 3. ∫
b
a
F(x)dx
预习自测
1. A 所求面积 S = ∫
2π
3
0
2sinxdx = - 2cosx |
2π
3
0
= - 2( - 1
2
- 1) = 3.
2. C 如果变速直线运动的速度为 v = v(t)(v(t) ≥ 0),那么从时刻 t =
a 到 t = b 所经过的路程是∫ba v(t)dt,
∴ ∫ t00 gtdt = 12 gt2 t00 =
1
2
g(t20 - 0) =
1
2
gt20 . 故应选 C.
3. 1
2
曲线 y = x2 与 y = cx3 的交点为 1
c
, 1
c2( ).
由题意知∫
1
c
0
(x2 - cx3 )dx = 1
3
x3 - c
4
x4( )
1
c
0 =
1
12c3
= 2
3
. ∴ c = 1
2
.
4. 4x + 3 ∵ f(x) 是一次函数,设 f(x) = ax + b(a ≠ 0),则∫10 f(x)dx =
∫10 (ax + b)dx = ∫
1
0
axdx + ∫10 bdx = 12 a +