微乙小考四 (2018/11/29)
1. (a) (2 分) 求 R sin(x)dx.
(b) (2 分) 求 R
01 1+x1 2dx
(c) (2 分) 若 f(x) = R
1xtan
10(2t)dt, 則 f
0(x) 為何?
sol: (a) R sin xdx = − cos x + C (b) R
10 1
1+x2
dx = tan
−1(x)|
10= tan
−1(1) − tan
−1(0) =
π4(c) f
0(x) = tan
10(2x)
2. (a) (4 分) 證明
654≤ R
4 01
1+x3
dx ≤ 4.
(b) (3 分) 假設 [0, 4] = [0, 2] ∪ [2, 4] 是 [0, 4] 的一個分割. 試求在此分割下, R
04 1+x13的最小黎曼和.
sol: (a) let f (x) =
1+x1 3is decreasing on [0,4]
so,
1+413=
651≤ f (x) ≤
1+01 3= 1 R
40 1 65
≤ R
40
f (x)dx ≤ R
4 01dx
4 65
≤ R
40 1 1+x3
≤ 4 (b) t
1= 2, t
2= 4
2
X
i=1
f (t
i) = f (2) · 2 + f (4) · 2 = 2 · ( 1
1 + 8 + 1
1 + 64 ) = 148 585 3. (a) (2 分) 陳述微積分基本定理.
(b) (3 分) 試求
dxdR
b(x)a(x)
f (t)dt . (c) (2 分) 使用 (b) 計算
dxdR
x3x2
ln(cos t)dt . sol: (a) if f is continuous on [a,b]
Let F (x) = R
xa
f (t)dt then F
0(x) = f (x) if G
0(x) = f (x) then R
ba
f (x)dx = G(b) − G(a) (b)
dxd( R
b(x)a(x)
f (t)dt) if f(t) is continuous on [c,d]
dxd( R
b(x)a(x)
f (t)dt) =
dxd( R
b(x)C
f (t)dt− R
a(x)C
f (t)dt)
=
dxd( R
b(x)C
f (t)dt) −
dxd( R
a(x)C
f (t)dt)
=
d(b(x))d( R
b(x)C
f (t)dt) ·
d(b(x))dx−
d(a(x))d( R
a(x)C
f (t)dt) ·
d(a(x))dx(by F.T.O.C)=f (b(x)) · b
0(x) − f (a(x)) · a
0(x)
(c) by(b)
d dx (
Z
x3 x2ln(cos t)dt) = ln(cos(x
3)) · 3x
2− ln(cos(x
2)) · 2x
1