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```Exercise in Mathematics I – End-of-term Exam (8 January 2008)
Yoshihito Yasaki

（※参考）
limit: 極限値
indeterminate form: 不定形
approximate: 近似する
to the third decimal place: 小数第 3 位まで
definite integral: 定積分
area: 面積
enclose: 囲う
x-axis: x 軸
improper integral: 広義積分
exist: 存在する
1. Find the limit lim
x →∞
lim
x →∞
(
(x
= lim
x →∞
)
x 2 + x + 1 − x 2 − 3x + 1
x →∞
= lim
)
)(
x2 + x + 1 + x2 − 3x + 1
(1) Approximate y = (1 + x)5 as a quadratic in x .
)
x 2 + x + 1 + x 2 − 3x + 1
2
+ x + 1) − ( x 2 − 3x + 1)
x 2 + x + 1 + x 2 − 3x + 1
4
1 1
3 1
1+ + 2 + 1− + 2
x x
x x
= lim
x →∞
=
4x
x 2 + x + 1 + x 2 − 3x + 1
4
=2
1+1
2. Find the limit of the indeterminate form
xe x − sin x
lim
.
x →0
x2
lim
x →0
( xe x − sin x )′ = lim e x + xe x − cos x
xe x − sin x
=
lim
x →0
x →0
x2
2x
( x 2 )′
(e
= lim
x
+ xe x − cos x )′
( 2 x )′
x →0
=
e x + e x + xe x + sin x
x →0
2
= lim
1+1+ 0 + 0
2
−1
x
with respect to x .
Taking logarithms of both sides, log y = sin −1 x .
y′
1
Differentiating with respect to x ,
=
y
1 − x2
y′ =
y
1 − x2
=
(2) Using the result from (1), approximate
the third decimal place.
5
(1)
esin
−1
x
1 − x2
1.015 to
y (0) = 1
y = (1 + x) 2
3
2
5
y ′ = (1 + x)
2
1
15
y ′′ = (1 + x) 2
4
5
2
15
y ′′(0) =
4
y ′(0) =
5
15
y = 1 + 2 x + 4 x 2 + ...
1!
2!
5
15
≈ 1 + x + x2
2
8
(2) Let x = 0.01 in the above approximation.
5
15
1.015 ≈ 1 + × 0.01 + × 0.0001 ≈ 1 + 0.025 + 0.0002
2
8
≈ 1.025
5. Find the indefinite integral
log x
∫ x(2 + log x) dx .
dt 1
= , so dx = x dt .
dx x
log x
t−2
⎛ 2⎞
∫ x(2 + log x) dx = ∫ xt x dt = ∫ ⎜⎝1 − t ⎟⎠ dt
= t − 2log t + C1 = 2 + log x − 2log 2 + log x + C1
Let t = 2 + log x .
3. Differentiate y = e sin
indefinite integral: 不定積分
region: 領域
curve: 曲線
evaluate: 求める
x 2 + x + 1 − x 2 − 3x + 1 .
x 2 + x + 1 − x 2 − 3x + 1
(
= lim
x →∞
(
differentiate: 微分する
Then,
= log x − 2log 2 + log x + C
∫ tan
6. Find the indefinite integral
−1
3 x dx .
（※ ∫ f ′gdx = fg − ∫ fg ′dx で f ′ = 1 ， g = tan −1 3 x ）
∫ sin
−1
3x
dx
1 + 9 x2
1
= x tan −1 3x − log (1 + 9 x 2 ) + C
6
3xdx = x tan −1 3x − ∫
9. Find the area S of the region enclosed by the
curve y = x 3 − x 2 − 2 x and the x-axis.
y = x 3 − x 2 − 2 x = x( x + 1)( x − 2)
The curve crosses the x-axis when x = −1, 0, 2 .
y
S1
−1
x
O
S = S1 + S 2 = ∫
5x − 3
∫ ( x + 1)( x − 3) dx .
7. Find the indefinite integral
0
−1
5x − 3
A
B
=
+
( x + 1)( x − 3) x + 1 x − 3
5 x − 3 = A( x − 3) + B ( x + 1)
When x = −1 , −8 = −4A , so A = 2 .
When x = 3 , 12 = 4B , so B = 3 .
5x − 3
3 ⎞
⎛ 2
∫ ( x + 1)( x − 3) dx = ∫ ⎜⎝ x + 1 + x − 3 ⎟⎠ dx
= 2log x + 1 + 3log x − 3 + C
(x
2
S2
3
− x 2 − 2 x ) dx − ∫ ( x 3 − x 2 − 2 x ) dx
2
0
0
2
⎡ x 4 x3
⎤
⎡ x 4 x3
⎤
= ⎢ − − x2 ⎥ − ⎢ − − x2 ⎥
3
3
⎣4
⎦ −1 ⎣ 4
⎦0
8
37
⎛1 1 ⎞ ⎛
⎞
= 0 − ⎜ + − 1⎟ − ⎜ 4 − − 4 ⎟ + 0 =
3
12
⎝4 3 ⎠ ⎝
⎠
Let
10. Does the improper integral
∫
∞
0
e − 2 x dx exist?
Evaluate it if it does.
K
⎡ 1
⎤
e
dx = lim ⎢ − e−2 x ⎥
∫ 0 e dx = Klim
→∞ ∫ 0
K →∞
⎣ 2
⎦0
1⎞ 1
⎛ 1
= lim ⎜ − e −2 K + ⎟ =
K →∞
2⎠ 2
⎝ 2
The improper integral exists.
∞
8. Find the definite integral
∫
3
0
dx
18 − x 2
Let x = 3 2 sin t
dx
= 3 2 cos t
dt
x = 0 ↔ sin t = 0 ↔ t = 0
1
π
x = 3 ↔ sin t =
↔t=
4
2
∫
3
0
dx
18 − x 2
π
=∫4
0
3 2 cos t dx
3 2 cos t
= [t ]0 =
π 4
π
4
.
−2 x
K
−2 x
```