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Solutions to various integration and limits problems. It includes the use of l'hopital's rule, integration by parts, and finding the length of a curve. The problems involve calculating limits of functions, integrating expressions, and finding the area under the curve of a function.
Typology: Exams
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Solution: 由定理4.3(l’Hopital 法則):
xlim→ 0
sin x − x cos x tan−^1 x − sin x = lim x→ 0 cos x − cos x + x sin x ( (^) 1+^1 x 2 ) − cos x = lim x→ 0 sin x + x cos x − 2 x (1+x^2 )^2 + sin^ x^
= lim x→ 0
(sinx^ x) + cos x − 2 (1+x^2 )^2 + ( sin x x )
(1)使用l’Hopital法則:第一次微對 3 分，第二次對 3 分，接下來的運算及答案 4 分。 (2)使用泰勒展式:一個展式對 3 分共 9 分，接下來的運算及答案 1 分。
x^2 + 3x + 2 dx.
Solution:
dx x^2 + 3x + 2
x + 1
x + 2
dx[5%] = ln
∣∣^ x^ + 1 x + 2
∣∣+C[5%] = ln |x+1|−ln |x+2|+C.
Note.
Solution: The area is equal to the following integration. ∫ (^) π
0
| sin x − cos x|dx
Because cos x ≥ sin x if 0 ≤ x ≤ π 4 and sin x ≥ cos x if π 4 ≤ x ≤ π, the area is equal to ∫ π 4
0
(cos x − sin x)dx +
∫ (^) π π 4 (sin x − cos x)dx (1)
Then ∫ (^) π
0
| sin x − cos x|dx =
∫ π 4
0
(cos x − sin x)dx +
∫ (^) π π 4 (sin x − cos x)dx
= (sin x + cos x) |
π 4 0 +(−^ cos^ x^ −^ sin^ x)^ |π^ π 4 = (
Grading evaluation:
x^2 由 x = 0 到 x = 1 的曲線長度.
Solution: f ′(x) = x length l =
0
1 + x^2 dx (7 pts)
∫ π 4
0
1 + tan^2 θ sec^2 θdθ (if we let x = tan θ,θ ∈ (− π 2
π 2
), then dx = sec^2 θdθ)
∫ π 4
0
sec^3 θdθ (3 pts)
∫ π 4
0
sec θd tan θ
= sec θ tan θ|
π 4 0 −
∫ π 4
0
sec θ tan^2 θdθ
= sec θ tan θ|
π 4 0 −
∫ π 4
0
sec θ(sec^2 θ − 1)dθ
= sec θ tan θ|
π 4 0 −
∫ π 4
0
sec^3 θdθ +
∫ π 4
0
sec θdθ
= (sec θ tan θ + ln | sec θ + tan θ|)|
π 4 0 −
∫ π 4
0
sec^3 θdθ
⇒ l =
∫ π 4
0
sec^3 θdθ =
(sec θ tan θ + ln | sec θ + tan θ|)|
π 4 0
=
ln (
(5 pts)
tan x x , y =
π 及 x = 0 所圍成的區域繞 y 軸旋轉之體積.
Solution: We have the following simple observations:
(i) lim x→ 0 tan x x
(ii)
tan x x
2 x − sin(2x) 2 x^2 cos^2 x
0 , ∀x ∈ (0, π 2
tan x x ↗ on (0, π 2
(iii) From (ii), tan x x
π has a unique solution which is x = π 4 clearly.
(iv ) tan x x
π on (0, π 4
Hence,
Area = 2 π
∫ (^) π/ 4
0
x
π
tan x x
dx
= 2 π
∫ (^) π/ 4
0
π x − tan x
dx
= 2 π
π
x^2
]π/ 4 0
ln | cos x|
]π/ 4 0 = π^2 4 − π ln 2 (2)
= π^2 4
Grade:
Remark. You don’t have to claim that x = π 4 is the unique solution to tan x x
π on (0, π 2
(a) (10%) 求 f (x) 對 x = 0 的泰勒展開式, 並寫出一般項.
(b) (10%) 以(a)中非零的前三項估計積分
0
f (x) dx, 誤差忽略不計.
Solution:
a. We know
ex^ =
0
xn n!
Therefore,
e−x^2 =
0
(−x^2 )n n!
b. ∫ (^12)
0
e−x^2 dx ≈
0
(1 − x^2 + x^4 2! )dx
=
