topikal matematik tambahan-pengamiran
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7/25/2019 Topikal Matematik Tambahan-Pengamiran
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1449/1 2015 Maths Catch Network © www.maths-catch.com [Lihat halaman sebelah]
SULIT
USAHA +DOA+TAWAKALMATHS Catch
PENGAMIRAN (INTEGRATION)1
Find ∫ −4
5 dx.
Cari ∫ −4
5 dx.
[1 mark ][1 markah]
Answer: Jawapan:
∫ −4
5 dx = −
4
5 x + c
2 Find ∫ −12 dx.Cari ∫ −12 dx.
[1 mark ][1 markah]
Answer: Jawapan:
∫ −12 dx = −12 x + c
3 Find ∫ x7 dx.Cari ∫ x7 dx.
[2 marks][2 markah]
Answer: Jawapan:
∫ x7 dx
= x7 + 1
7 + 1 c
=1
8 x8 + c
4 Find ∫ −
17
8 x4 dx.
Cari ∫ −17
8 x4 dx.
[2 marks]
[2 markah]Answer:
Jawapan:
∫ −17
8 x4 dx
= −17
8 ( x4 + 1
4 + 1) + c
= −17
40 x5 + c
5 Find ∫ (
19
10 x4 +
3
13 x11) dx.
Cari ∫ ( 19
10 x4 +
3
13 x11 ) dx.
[2 marks][2 markah]
Answer:
Jawapan:
∫ (19
10 x4 +
3
13 x11) dx
=19
10 ( x4 + 1
4 + 1) +
3
13 (
x11 + 1
11 + 1) + c
=19
50 x5 +
1
52 x12 + c
6 Find (−6 x − 2) dx.
Cari ∫ ( −6x − 2)2 dx.
[2 marks][2 markah]
Answer: Jawapan:
∫ (−6 x − 2)2 dx = ∫ (36 x2 + 24 x + 4) dx = 12 x3 + 12 x2 + 4 x + c
7 Find ∫ (2 x3
+ 6 x2
) dx.
Cari ∫ (2x
1
3
+ 6x
5
2
) dx.
[2 marks][2 markah]
Answer: Jawapan:
∫ (2 x
1
3
+ 6 x
5
2
) dx
= 2(3
4 ) x
4
3
+ 6(2
7 ) x
7
2
+ c
=3
2 x
4
3
+12
7 x
7
2
+ c
8 Given the gradient function
dy
dx = 2 x−25, find the equation
of the curve.
Diberi fungsi kecerunandy
dx = 2x−25 , cari persamaan
lengkung itu.
[2 marks][2 markah]Answer:
Jawapan:
dy
dx = 2 x−25
∴ y = ∫ 2 x−25 dx
=2 x−25 + 1
−25 + 1 c
= −1
12 x24 + c
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USAHA +DOA+TAWAKALMATHS Catch
9 Given the gradient function
dy
dx = −
3
16 x18, find the
equation of the curve.
Diberi fungsi kecerunandy
dx = −
3
16 x18 , cari persamaan
lengkung itu.
[2 marks]
[2 markah]Answer:
Jawapan:
dy
dx = −
3
16 x18
∴ y = ∫ −3
16 x18 dx
= −3
16 (
x18 + 1
18 + 1) + c
= −3
304 x19 + c
10 Given the gradient function
dy
dx = (
8
7 x−16 +
10
9 x16), find
the equation of the curve.
Diberi fungsi kecerunandy
dx = (
8
7 x−16 +
10
9 x16 ), cari
ersamaan lengkung itu. [2 marks]
[2 markah]Answer:
Jawapan:
dy
dx = (
8
7 x−16 +
10
9 x16)
∴ y = ∫ (8
7 x−16
+10
9 x16
) dx
=8
7 (
x−16 + 1
−16 + 1) +
10
9 (
x16 + 1
16 + 1) + c
= −8
105 x15 +10
153 x17 + c
11 Given the gradient function
dy
dx = (3 x + 1)2, find the
equation of the curve.
Diberi fungsi kecerunandy
dx = (3x + 1)2 , cari persamaan
lengkung itu.
[2 marks]
[2 markah]
Answer: Jawapan:
dy
dx = (3 x + 1)2
∴ y = ∫ (3 x + 1)2 dx
= ∫ (9 x2 + 6 x + 1) dx = 3 x3 + 3 x2 + x + c
12 Given the gradient function
dy
dx = (−4 x 4
+ 9 x−
11
), find the
equation of the curve.
Diberi fungsi kecerunan dydx
= ( −4x
11
4
+ 9x−
8
11
), cari
ersamaan lengkung itu.
[2 marks][2 markah]
Answer:
Jawapan:
dy
dx = (−4 x
11
4
+ 9 x−
8
11
)
∴ y = ∫ (−4 x
11
4
+ 9 x−
8
11
) dx
= −4(4
15 ) x
15
4
+ 9(11
3 ) x
3
11
+ c
= −16
15 x
15
4
+ 33 x
3
11
+ c
13 Given
dy
dx = (−5 x + 6)2, find the equation of the curve y =
( x) if the curve passes through the point (1, 12
1
3 ).
Diberidy
dx = ( −5x + 6)2 , cari persamaan lengkung y = f(x
ika lengkung itu melalui titik (1, 121
3 ).
[2 marks][2 markah]
Answer: Jawapan:
dy
dx = (−5 x + 6)2
y = ∫ (−5 x + 6)2 dx
y = ∫ (25 x2 − 60 x + 36) dx
=25
3 x3 − 30 x2 + 36 x + c
When x = 1, y =37
3
37
3 =
25
3 (1)3 − 30(1)2 + 36(1) + c
c = −2
∴ y =25
3 x3 − 30 x2 + 36 x − 2
14 Given
dy
dx = −3 x−2 − 6 x, find the equation of the curve y =
( x) if the curve passes through the point (1, 3).
Diberidy
dx = −3x−2 − 6x, cari persamaan lengkung y =
(x jika lengkung itu melalui titik (1, 3). [2 marks]
[2 markah]
Answer:
Jawapan:
dy
dx = −3 x−2
− 6 x
y = ∫ (−3 x−2 − 6 x) dx
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= −3( x−2 + 1
−2 + 1) − 6(
x1 + 1
1 + 1) + c
=3
x − 3 x2 + c
When x = 1, y = 3
3 =3
1(1) − 3(1)2 + c
c = 3
y =3 x
− 3 x2 + 3
15 Given that (−3 x + 2) dx = kx + 2 x + c, where k and c are constants, find
Diberi ∫ ( −3x2 + 2) dx = kx3 + 2x + c, dengan keadaan kdan c ialah pemalar, cari
(a) the value of k
nilai k
(b) the value of c if (−3 x + 2) dx = −1 when x = 1.
nilai c jika ∫ ( −3x2 + 2) dx = −1 apabila x = 1.
[3 marks][3 markah]
Answer:
Jawapan:
(a) (−3 x + 2) dx
= − x3 + 2 x + c
∴ k = −1
(b) When x = 1,
∫ (−3 x2 + 2) dx = −1
−1(1)3 + 2(1) + c = −1c = −2
16 The variables x and y are related by the equation y = x−3, where p is a constant.
Pemboleh ubah x dan y dihubungkan oleh persamaan y
= px−3 , dengan keadaan p ialah pemalar.
(a) Convert the equation y = px−3 to linear form.Tukarkan persamaan y = px−3 kepada bentuk linear.
(b) Diagram 1 shows the straight line obtained by
plotting log10 y against log10 x.
Rajah 1 menunjukkan graf garis lurus yangdiperoleh dengan memplot log 10 y melawan log 10 x.
Diagram 1 Rajah 1
Find the value ofCarikan nilai
(i) log10 p.
(ii) q.
[4 marks][4 markah]
Answer: Jawapan:
(a) log10 y = log10 p − 3(log10 x)= −3(log10 x) + log10 p
(b) (i) log10 p = Y −intercept
= 3
(ii) log10 y = −3(2) + 3= −3
q = log10 y = −3
17 Find ∫ (5 x + 2)9
dx.
Cari ∫ (5x + 2)
4
9
dx.
[2 marks][2 markah]
Answer: Jawapan:
∫ (5 x + 2)
4
9
dx
=
(5 x + 2)4
9 + 1
5(4
9 + 1)
+ c
=9
65 (5 x + 2)
13
9
+ c
18 Find ∫
17
(12 x − 18)12 dx.
Cari ∫ 17
(12x − 18)12 dx.
[2 marks][2 markah]
Answer: Jawapan:
Let u = 12 x − 18du
dx = 12
dx =1
12 du
∫ 17(12 x − 18)12 dx
= ∫ 17
(u)12 (1
12 du)
= ∫ 17
12 u−12 du
= −17
132 u−11 + c
= − 17
132u11 + c
= − 17
132(12 x − 18)11 + c
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USAHA +DOA+TAWAKALMATHS Catch
19 Given that ∫ 4 x3 dx = 20 when x = 2. Find the value of theconstant c.
Diberi ∫ 4x3 dx = 20 apabila x = 2. Cari nilai pemalar c.
[2 marks][2 markah]
Answer:
Jawapan:
∫ 4 x3 dx = x4 + c
When x = 2,
x4 + c = 20(2)4 + c = 20c = 4
20 Given that ∫ (4 x − 5)2 dx =
50
3 when x = 2. Find the value
of the constant c.
Diberi ∫ (4x − 5)2 dx =50
3 apabila x = 2. Cari nilai
emalar c.
[2 marks]
[2 markah]Answer: Jawapan:
∫ (4 x − 5)2 dx =50
3
y = ∫ (4 x − 5)2 dx
y = ∫ (16 x2 − 40 x + 25) dx
=16
3 x3 − 20 x2 + 25 x + c
When x = 2, y =50
3
50
3 =
16
3 (2)3 − 20(2)2 + 25(2) + c
c = 4
21 Given
d
dx (5 x2 − x − 6) = f ( x), find ∫
6
4 f ( x) dx.
Diberid
dx (5x2 − x − 6) = f(x), cari ∫
6
4 f(x) dx.
[2 marks]
[2 markah]Answer:
Jawapan:
d
dx (5 x2 − x − 6) = f ( x) = [5 x2 − x − 6]
6
4
= 5(6)2 − (6) − 6 − [5(4)2 − (4) − 6]= 168 − (70)
= 98
22 Given that ∫
9
7 f ( x) dx = 2 and ∫
9
7 [7 f ( x) + kx] dx = 110, find
the value of k .
Diberi ∫9
7 f(x) dx = 2 dan ∫
9
7 [7f(x) + kx] dx = 110,
carikan nilai k.[2 marks]
[2 markah]
Answer:
Jawapan:
∫9
7 [7 f ( x) + kx] dx = 110
7∫9
7 f ( x) dx + ∫
9
7 kx dx = 110
7(2) + k [
x2
2 ]
9
7 = 110
14 + k (81
2 −
49
2 ) = 110
k = 6
23 Given that ∫
4
3 f ( x) dx = 1, find
Diberi ∫4
3 f(x) dx = 1, carikan
(a)the value of ∫
3
4 f ( x) dx.
niali ∫3
4 f(x) dx.
(b)the value of k if ∫
4
3 [kx − 5 f ( x)] dx =
39
2 .
nilai k jika ∫43
[kx − 5f(x)] dx =392
.
[4 marks][4 markah]
Answer: Jawapan:
(a)∫3
4 f ( x) dx = −∫
4
3 f ( x) dx
= −1
(b)∫4
3 [kx − 5 f ( x)] dx =
39
2
∫4
3 kx dx − 5∫
4
3 f ( x) dx =
39
2
k [ x2
2 ]
4
3 − 5(1) =
39
2
k (8 − 9
2 ) − 5 =
39
2
k = 7
24 Given that ∫
5
7 f ( x) dx = −9, find
Diberi ∫5
7 f(x) dx = −9, carikan
(a)∫7
5
f ( x) dx
(b)∫5
7 [2 + 3 f ( x)] dx
[4 marks][4 markah]
Answer:
Jawapan:
(a)∫7
5 f ( x) dx = −∫
5
7 f ( x) dx
= 9
(b)∫5
7 [2 + 3 f ( x)] dx
= ∫5
7 2 dx + 3∫
5
7 f ( x) dx
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= [2 x]5
7 + 3(−9)
= (10 − 14) − 27
= −31
25 Given that ∫
b
0 f ( x) dx = 1 and ∫
b
0 f ( x + 4) dx = 9, find the
value of b.
Diberi ∫
b
0 f(x) dx = 1 dan ∫
b
0 f(x + 4) dx = 9, carikan nilaib.
[2 marks]
[2 markah]Answer: Jawapan:
∫b
0 f ( x + 4) dx = 9
∫b
0 f ( x) dx + ∫
b
04 dx = 9
1 + [4 x]b
0 = 9
4(b) = 8
b = 2b = 2
26 Diagram 2 shows the curve y = f ( x) cutting the x-axis at x
= a and x = b. Rajah 2 menunjukkan lengkung y = f(x) yang memotong
aksi-x di x = a dan x = b.
Diagram 2 Rajah 2
Given that the area of the shaded region is 4 unit2, find
the value of ∫b
a 4 f ( x) dx.
Diberi luas rantau berlorek ialah 4 unit 2 , cari nilai ∫b
a
4f(x) dx. [2 marks]
[2 markah]Answer:
Jawapan:
∫b
a f ( x) dx = −4
∫b
a 4 f ( x) dx = 4∫
b
a f ( x) dx
= 4(−4)= −16
27 Diagram 3 shows the curve y = f ( x) cutting the x-axis at x
= k and passing the point (20, 20).
Rajah 3 menunjukkan lengkung y = f(x) yang memotongaksi-x di x = k dan melalui titik (20, 20).
Diagram 3 Rajah 3
Given that the area of the shaded region is 280 unit2, find
the value of ∫20
k f ( x) dx.
Diberi luas rantau berlorek ialah 280 unit 2 , cari nilai ∫20
k
(x) dx. [2 marks]
[2 markah]Answer: Jawapan:
20 × 20 − ∫20
k f ( x) dx = 280
∫20
k f ( x) dx = 20 × 20 − 280
= 120
28 Diagram 4 shows the curve y = f ( x) cutting the x-axis at x = 0, x = a and x = b.
Rajah 4 menunjukkan lengkung y = f(x) yang memotong
aksi-x di x = 0, x = a dan x = b.
Diagram 4 Rajah 4
Given that the area of the shaded region P is 5 unit2 andthe area of the shaded region Q is 4 unit.
Find the value of ∫a0 6 f ( x) dx + ∫ba 4 f ( x) dx.
Diberi luas rantau berlorek P ialah 5 unit 2 dan luasrantau berlorek Q ialah 4 unit.
Cari nilai ∫a
0 6f(x) dx + ∫
b
a 4f(x) dx.
[2 marks][2 markah]
Answer: Jawapan:
∫a
0
6 f ( x) dx + ∫b
a
4 f ( x) dx
= 6∫a
0 f ( x) dx + 4∫
b
a f ( x) dx
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= 6(5) + 4(−4)= 14
29 Diagram 5 shows the curve y = f ( x) cutting the x-axis at x
= b. Rajah 5 menunjukkan lengkung y = f(x) yang memotong
aksi-x di x = b.
Diagram 5 Rajah 5
Given that the area of the shaded region is 36 unit2, find
the value of ∫b
7 9 f ( x) dx.
Diberi luas rantau berlorek ialah 36 unit 2 , cari nilai ∫b
7
9f(x) dx.
[2 marks][2 markah]
Answer:
Jawapan:
∫b
7 f ( x) dx = 36 − 7 × 4
= 8
∫b
7 9 f ( x) dx = 9∫
b
7 f ( x) dx
= 9(8)= 72
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