ch.6 integration
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3. (a) Express in partial fractions.
(3)
(b) Hence find the exact value of , giving your answer as a single
logarithm.
(5)
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6
2
5 3d
(2 3)( 2)
xx
x x
+− +∫
5 3
(2 3)( 2)
x
x x
+− +
*N20232B0424*
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4. Use the substitution x = sin θ to find the exact value of
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32
12
20
1d
(1 ).x
x−∫
*N20232B0624*
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5. Figure 1
Figure 1 shows the graph of the curve with equation
y = xe2x, x 0.
The finite region R bounded by the lines x = 1, the x-axis and the curve is shown shaded
in Figure 1.
(a) Use integration to find the exact value for the area of R.
(5)
(b) Complete the table with the values of y corresponding to x = 0.4 and 0.8.
(1)
(c) Use the trapezium rule with all the values in the table to find an approximate value
for this area, giving your answer to 4 significant figures.
(4)
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*N20232B0824*
x 0 0.2 0.4 0.6 0.8 1
y = xe2x 0 0.29836 1.99207 7.38906
y
x0 0.2 0.4 0.6 0.8 1
R
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Question 5 continued
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Turn over*N20232B0924*
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2. (a) Given that y = sec x, complete the table with the values of y corresponding to x = , and .
(2)
(b) Use the trapezium rule, with all the values for y in the completed table, to obtain anestimate for . Show all the steps of your working, and give your answerto 4 decimal places.
(3)
The exact value of is .
(c) Calculate the % error in using the estimate you obtained in part (b).(2)
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ln(1 2)4
0sec dx x
4
0sec dx x
4816
*N23553A0420*
x 0 16 8316 4
y 1 1.20269
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3. Using the substitution u2 = 2x – 1, or otherwise, find the exact value of
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5
1
3 d .(2 1)
x xx
*N23553A0620*
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4. Figure 1
Figure 1 shows the finite shaded region, R, which is bounded by the curve y = xex, the linex = 1, the line x = 3 and the x-axis.
The region R is rotated through 360 degrees about the x-axis.
Use integration by parts to find an exact value for the volume of the solid generated.
(8)
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*N23553A0820*
y
O 1 3 x
y = xex
R
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3. Figure 1
The curve with equation , 0 x 2π, is shown in Figure 1. The finite region
enclosed by the curve and the x-axis is shaded.
(a) Find, by integration, the area of the shaded region.(3)
This region is rotated through 2π radians about the x-axis.
(b) Find the volume of the solid generated.(6)
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3sin2xy =
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y
O x2π
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6. Figure 3
Figure 3 shows a sketch of the curve with equation y = (x – 1) ln x, x > 0.
(a) Complete the table with the values of y corresponding to x = 1.5 and x = 2.5.
(1)Given that I =∫
3
1(x – 1)ln x dx ,
(b) use the trapezium rule
(i) with values of y at x = 1, 2 and 3 to find an approximate value for I to4 significant figures,
(ii) with values of y at x = 1, 1.5, 2, 2.5 and 3 to find another approximate value forI to 4 significant figures.
(5)
(c) Explain, with reference to Figure 3, why an increase in the number of valuesimproves the accuracy of the approximation.
(1)
(d) Show, by integration, that the exact value of ∫3
1(x – 1)ln x dx is .
(6)
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32 ln 3
*N23563A01420*
1O
y
x
x
y
1
0
1.5 2
ln 2
2.5 3
2 ln 3
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15
Question 6 continued
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Turn over*N23563A01520*
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2. Figure 1
The curve with equation , is shown in Figure 1.
The region bounded by the lines x = , x = , the x-axis and the curve is shown shadedin Figure 1.
This region is rotated through 360 degrees about the x-axis.
(a) Use calculus to find the exact value of the volume of the solid generated.(5)
Figure 2
Figure 2 shows a paperweight with axis of symmetry AB where AB = 3 cm. A is a pointon the top surface of the paperweight, and B is a point on the base of the paperweight.The paperweight is geometrically similar to the solid in part (a).
(b) Find the volume of this paperweight.(2)
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12
14
12
1 ,3(1 2 )
y xx
*N23562A0420*
y
B
A
14
12
x0
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Question 2 continued
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Turn over
Q2
(Total 7 marks)
*N23562A0520*
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8. .
(a) Given that y = e (3x + 1), complete the table with the values of y corresponding to x = 2,3 and 4.
(2)
(b) Use the trapezium rule, with all the values of y in the completed table, to obtain anestimate for the original integral I, giving your answer to 4 significant figures.
(3)
(c) Use the substitution t = (3x + 1) to show that I may be expressed as , givingthe values of a, b and k.
(5)
(d) Use integration by parts to evaluate this integral, and hence find the value of I correctto 4 significant figures, showing all the steps in your working.
(5)
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e dtb
akt t
(3 + 1)5
0e dxI x
*N23562A01820*
x
y
0
e1
1
e2
2 3 4 5
e4
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Question 8 continued
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TOTAL FOR PAPER: 75 MARKS
END
Q8
(Total 15 marks)
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2. Use the substitution u = to find the exact value of
(6)
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1
20
2(2 1)
x
x dx+∫ .
2x
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*N26110A0624*
3. (a) Find . (4)
(b) Hence, using the identity , deduce 2cosx x dx∫ .(3)
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cos 2x x dx∫
2cos 2 2cos 1x x= −
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*N26110A0824*
4.
(a) Find the values of the constants A, B and C.(4)
(b) Hence show that the exact value of , giving the
value of the constant k.(6)
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22( 4 1 ) .(2 1 )(2 1 ) (2 1 ) (2 1 )
x B C Ax x x x
+ ≡ + +
+ − + −
2 2
1
2(4 1)(2 1)(2 1)
x dxx x
++ −∫ is 2 + ln k
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*N26110A01624*
7.
Figure 1
Figure 1 shows part of the curve with equation . The finite region R, which is bounded by the curve, the x-axis and the line , is shown shaded in
Figure 1.
(a) Given that , complete the table with the values of y corresponding to
, giving your answers to 5 decimal places.
x 0
y 0 1
(3)
(b) Use the trapezium rule with all the values of y in the completed table to obtain an estimate for the area of the shaded region R, giving your answer to 4 decimal places.
(4)
The region R is rotated through 2 radians around the x-axis to generate a solid of revolution.
(c) Use integration to find an exact value for the volume of the solid generated.(4)
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y
xO π4
R
y = (tan x)
x = 4π
x =16 16π and, 8
π π3
16π
π
y = (tan x)
8π
16π3
4π
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Turn over*N26110A01724*
Question 7 continued
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*N26282A0224*
1.
Figure 1
The curve shown in Figure 1 has equation , . The finite region Rbounded by the curve and the x-axis is shown shaded in Figure 1.
(a) Complete the table below with the values of y corresponding to x = and , giving your answers to 5 decimal places.
(2)
(b) Use the trapezium rule, with all the values in the completed table, to obtain an estimate for the area of the region R. Give your answer to 4 decimal places.
(4)
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x 0
y 0 8.87207 0
R
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*N26282A0624*
3.
Figure 2
The curve shown in Figure 2 has equation . The finite region bounded by the
curve, the x-axis and the lines x = a and x = b is shown shaded in Figure 2. This region is rotated through 360° about the x-axis to generate a solid of revolution.
Find the volume of the solid generated. Express your answer as a single simplified fraction, in terms of a and b.
(5)
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4. (i) Find .(4)
(ii) Find the exact value of .
(5)
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sin2
4
2 x xdπ
π
∫
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*H30427A0228*
1.
Figure 1
Figure 1 shows part of the curve with equation y x= e0 5 2. . The finite region R, shown shaded in Figure 1, is bounded by the curve, the x-axis, the y-axis and the line x = 2.
(a) Complete the table with the values of y corresponding to x = 0.8 and x = 1.6.
x 0 0.4 0.8 1.2 1.6 2
y e0 e0.08 e0.72 e2
(1)
(b) Use the trapezium rule with all the values in the table to find an approximate value for the area of R, giving your answer to 4 significant figures.
(3)
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y
x
1
O 2
R
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*H30427A0428*
2. (a) Use integration by parts to find .(3)
(b) Hence find . .(3)
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x xxe d∫x xx2 e d∫
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*N31013A0428*
2.
Figure 1
Figure 1 shows part of the curve The region R is bounded by the curve,
the x-axis, and the lines x = 0 and x = 2, as shown shaded in Figure 1.
(a) Use integration to find the area of R.(4)
The region R is rotated 360° about the x-axis.
(b) Use integration to find the exact value of the volume of the solid formed.(5)
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R
y
3
O 2 x
yx
31 4( )
.
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*N31013A0528* Turn over
Question 2 continued
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*N31013A02028*
6. (a) Find .(2)
(b) Use integration by parts to find
(4)
(c) Use the substitution u = 1 + ex to show that
where k is a constant.(7)
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tan2 x xd∫1
3xx xln .∫ d
ee
d e e e3
2
111
2
x
xx x xx k
+= − + + +∫ ln( ) ,
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*N31013A02128* Turn over
Question 6 continued
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*H34265A0428*
2.
Figure 1
Figure 1 shows the finite region R bounded by the x-axis, the y-axis and the curve with
equation y = 3 cos x3
, 0 x 32
.
The table shows corresponding values of x and y for y = 3 cos x3
.
x 038
34
98
32
y 3 2.77164 2.12132 0
(a) Complete the table above giving the missing value of y to 5 decimal places. (1)
(b) Using the trapezium rule, with all the values of y from the completed table, find an approximation for the area of R, giving your answer to 3 decimal places.
(4)
(c) Use integration to find the exact area of R. (3)
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3
O x
y
R
32π
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*H34265A0528* Turn over
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___________________________________________________________________________ Q2
(Total 8 marks)
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*H34265A0628*
3. f(x xx x x
Ax
Bx
Cx
)( )( )( )
=−
+ + +=
++
++
+4 2
2 1 1 3 2 1 1 3
(a) Find the values of the constants A, B and C.(4)
(b) (i) Hence find f d( ) .x x(3)
(ii) Find f d( )x x0
2 in the form 1n k, where k is a constant.
(3)
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6. (a) Find √ ( ) .5−∫ x xd(2)
Figure 3
Figure 3 shows a sketch of the curve with equation
y = (x – 1) (5 – x), 1 x 5
(b) (i) Using integration by parts, or otherwise, find
( ) ( )x x x− −∫ 1 5√ d(4)
(ii) Hence find ( ) ( )x x x− −∫ 1 51
5√ d .
(2)
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O x
y
1 5
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8. (a) Using the identity cos 2 = 1 – 2 sin2 , find .(2)
Figure 4
Figure 4 shows part of the curve C with parametric equations
x = tan , y = 2 sin 2 , 0 <2
The finite shaded region S shown in Figure 4 is bounded by C, the line x = 1
and the
x-axis. This shaded region is rotated through 2 radians about the x-axis to form a solid of revolution.
(b) Show that the volume of the solid of revolution formed is given by the integral
k sin2
0
6π
θ θ d∫ where k is a constant.
(5)
(c) Hence find the exact value for this volume, giving your answer in the form p 2 + q 3, where p and q are constants.
(3)
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O x
y
1√ 3
C
S
sin2∫ θ θd
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Question 8 continued
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TOTAL FOR PAPER: 75 MARKS
END
Q8
(Total 10 marks)
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2.
Figure 1
Figure 1 shows a sketch of the curve with equation y = x1n x, x 1. The finite region R,shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 4.
The table shows corresponding values of x and y for y = x1n x.
x 1 1.5 2 2.5 3 3.5 4
y 0 0.608 3.296 4.385 5.545
(a) Complete the table with the values of y corresponding to x = 2 and x = 2.5, giving your answers to 3 decimal places.
(2)
(b) Use the trapezium rule, with all the values of y in the completed table, to obtain an estimate for the area of R, giving your answer to 2 decimal places.
(4)
(c) (i) Use integration by parts to find ln d .x x x∫ (ii) Hence find the exact area of R, giving your answer in the form (1 ln 2 ,
4a b)+
where a and b are integers.(7)
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O x41
y
R
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8. (a) Using the substitution x = 2 cos u, or otherwise, find the exact value of
142 2
1 x xx
( )d
2
(7)
Figure 3
Figure 3 shows a sketch of part of the curve with equation yx x
x4
40 2
214( )
, .
The shaded region S, shown in Figure 3, is bounded by the curve, the x-axis and the lines with equations x = 1 and x = 2. The shaded region S is rotated through 2 radians about the x-axis to form a solid of revolution.
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed.
(3)
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y
S
O 1 x√2
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Question 8 continued
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TOTAL FOR PAPER: 75 MARKSEND
Q8
(Total 10 marks)
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1.
Figure 1
Figure 1 shows part of the curve with equation 2(0.75 cos ).y x= + The finite region R, shown shaded in Figure 1, is bounded by the curve, the y-axis, the x-axis and the line with equation x =
π3
.
(a) Complete the table with values of y corresponding to x=π6
and x=π4
.
x 0 12π π
6π4
π3
y 1.3229 1.2973 1
(2)
(b) Use the trapezium rule
(i) with the values of y at 0,x = x=π6
and x=π3
to find an estimate of the area of R.
Give your answer to 3 decimal places.
(ii) with the values of y at 0,x = x=12π , x =
π6
, x =π4
and x =π3
to find a
further estimate of the area of R. Give your answer to 3 decimal places.(6)
O
y
x
R
3π
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2. Using the substitution cos 1,u x= + or otherwise, show that
2 cos 1
0
e sin d e(e 1)x x xπ
+ = −∫(6)
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*H35405A0224*
1. Use integration to find the exact value of
x x xsin 20
2 d
π
∫(6)
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6. The curve C has parametric equations
lnx t= , 2 2y t= − , 0t
Find
(a) an equation of the normal to C at the point where 3t = ,(6)
(b) a cartesian equation of C.(3)
O
y
x
C
R
ln 2 ln 4
Figure 1
The finite area R, shown in Figure 1, is bounded by C, the x-axis, the line ln 2x = and the line ln 4x = . The area R is rotated through360° about the x-axis.
(c) Use calculus to find the exact volume of the solid generated.(6)
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7.
(a) Given that y
x=
+ −1
4 1( ) , complete the table below with values of y corresponding
to 3x = and 5x = . Give your values to 4 decimal places.
x 2 3 4 5
y 0.2 0.1745
(2)
(b) Use the trapezium rule, with all of the values of y in the completed table, to obtain an estimate of I, giving your answer to 3 decimal places.
(4)
(c) Using the substitution ( )24 1x u= − + , or otherwise, and integrating, find the exact value of I.
(8)
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Ix
x=+ −∫ 1
4 12
5
( )d
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Question 7 continued
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TOTAL FOR PAPER: 75 MARKS
END
Q7
(Total 14 marks)
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4.
O x√2
y
R
Figure 2
Figure 2 shows a sketch of the curve with equation y = x3 ln (x2 + 2), x 0. The finite region R, shown shaded in Figure 2, is bounded by the curve, the x-axis and the
line 2.x =
The table below shows corresponding values of x and y for y = x3 ln (x2 + 2).
x 0 2
42
2
3 2
42
y 0 0.3240 3.9210
(a) Complete the table above giving the missing values of y to 4 decimal places.(2)
(b) Use the trapezium rule, with all the values of y in the completed table, to obtain an estimate for the area of R, giving your answer to 2 decimal places.
(3)
(c) Use the substitution u = x2 + 2 to show that the area of R is
4
2
1( 2) ln d
2−∫ u u u
(4)
(d) Hence, or otherwise, find the exact area of R.(6)
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7.y
xO
l C
P
S
Q√3
Figure 3
Figure 3 shows part of the curve C with parametric equations
x = tan , y = sin , 0 < 2
The point P lies on C and has coordinates 1
23, 3⎛ ⎞
⎜ ⎟⎝ ⎠.
(a) Find the value of at the point P.(2)
The line l is a normal to C at P. The normal cuts the x-axis at the point Q.
(b) Show that Q has coordinates ( )3, 0 ,k giving the value of the constant k.(6)
The finite shaded region S shown in Figure 3 is bounded by the curve C, the line 3x = and the x-axis. This shaded region is rotated through 2 radians about the x-axis to form a solid of revolution.
(c) Find the volume of the solid of revolution, giving your answer in the form p 3 + q 2, where p and q are constants.
(7)
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2. (a) Use integration by parts to find x x xsin .3 d∫(3)
(b) Using your answer to part (a), find x x x2 3cos .d∫(3)
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4.
O 2
y
x
S
Figure 1
Figure 1 shows the curve with equation
y xx
x=+
⎛⎝⎜
⎞⎠⎟
23 4
02 , �
The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2
The region S is rotated 360° about the x-axis.
Use integration to find the exact value of the volume of the solid generated, giving your answer in the form k ln a, where k and a are constants.
(5)
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6.
O
R
x
y
2π
Figure 3
Figure 3 shows a sketch of the curve with equation y xx
x=+
2 21
02
sin( cos )
, . � � π
The finite region R, shown shaded in Figure 3, is bounded by the curve and the x-axis.
The table below shows corresponding values of x and y for y xx
=+
2 21
sin( cos )
.
x 0π8
π4
38π π
2
y 0 1.17157 1.02280 0
(a) Complete the table above giving the missing value of y to 5 decimal places. (1)
(b) Use the trapezium rule, with all the values of y in the completed table, to obtain an estimate for the area of R, giving your answer to 4 decimal places.
(3)
(c) Using the substitution u = 1 + cos x , or otherwise, show that
2 21
4 1 4sin( cos )
ln( cos ) cosxx
x x x k+
= + − +∫ d
where k is a constant.(5)
(d) Hence calculate the error of the estimate in part (b), giving your answer to 2 significant figures.
(3)
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*P41484A0232*
1. f ( )( ) ( ) ( )
xx x
Ax
Bx
Cx
=−
= +−
+−
13 1 3 1 3 12 2
(a) Find the values of the constants A, B and C.(4)
(b) (i) Hence find f d( ) .x x∫ (ii) Find f d( )x x
1
2
∫ , leaving your answer in the form a b+ ln ,
where a and b are constants. (6)
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*P41484A02432*
7.
O x41
y
R
Figure 3
Figure 3 shows a sketch of part of the curve with equation y x x=12 2ln .
The finite region R, shown shaded in Figure 3, is bounded by the curve, the x-axis and the lines x = 1 and x = 4
(a) Use the trapezium rule, with 3 strips of equal width, to find an estimate for the area of R, giving your answer to 2 decimal places.
(4)
(b) Find x x x12 2ln .d∫
(4)
(c) Hence find the exact area of R, giving your answer in the form a bln ,2 + where a and b are exact constants.
(3)
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4
*P41860A0428*
2. (a) Use integration to find
13x∫ ln x dx
(5)
(b) Hence calculate
13
1
2
x∫ ln x dx
(2)
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4.
Figure 1
Figure 1 shows a sketch of part of the curve with equation y xx
=+1 √
. The finite region
R, shown shaded in Figure 1, is bounded by the curve, the x-axis, the line with equation x = 1 and the line with equation x = 4.
(a) Complete the table with the value of y corresponding to x = 3, giving your answer to 4 decimal places.
(1)
x 1 2 3 4
y 0.5 0.8284 1.3333
(b) Use the trapezium rule, with all the values of y in the completed table, to obtain an estimate of the area of the region R, giving your answer to 3 decimal places.
(3)
(c) Use the substitution u = 1 + �x, to find, by integrating, the exact area of R.(8)
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x
R
41O
y
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6.
Figure 3
Figure 3 shows a sketch of part of the curve with equation y = 1 – 2cos x, where x is measured in radians. The curve crosses the x-axis at the point A and at the point B.
(a) Find, in terms of �, the x coordinate of the point A and the x coordinate of the point B.(3)
The finite region S enclosed by the curve and the x-axis is shown shaded in Figure 3. The region S is rotated through 2� radians about the x-axis.
(b) Find, by integration, the exact value of the volume of the solid generated.(6)
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BAO
y
x
S
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*P43137A0232*
1. (a) Find x xx2e d .∫(5)
(b) Hence find the exact value of x xx2
0
1e d .∫
(2)
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3.
Figure 1
Figure 1 shows the finite region R bounded by the x-axis, the y-axis, the line 2πx = and
the curve with equation
1sec , 02 2
πy x x⎛ ⎞= ⎜ ⎟⎝ ⎠� �
The table shows corresponding values of x and y for y = sec 12
x⎛⎝⎜
⎞⎠⎟
.
x 0 6π
3π
2π
y 1 1.035276 1.414214
(a) Complete the table above giving the missing value of y to 6 decimal places.(1)
(b) Using the trapezium rule, with all of the values of y from the completed table, find an approximation for the area of R, giving your answer to 4 decimal places.
(3)
Region R is rotated through 2� radians about the x-axis.
(c) Use calculus to find the exact volume of the solid formed.(4)
y
y x= ⎛⎝⎜
⎞⎠⎟
sec 12
2π x
R
O
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5. (a) Use the substitution x = u2, u > 0, to show that
2d =(2 1)
duxu u −
1(2x x − 1)∫ ∫
(3)
(b) Hence show that9
1
1 d = 2ln axb
⎛ ⎞⎜ ⎟⎝ ⎠∫ (2x x − 1)
where a and b are integers to be determined.(7)
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Edexcel AS/A level Mathematics Formulae List: Core Mathematics C4 – Issue 1 – September 2009 7
Core Mathematics C4
Candidates sitting C4 may also require those formulae listed under Core Mathematics C1, C2 and C3.
Integration (+ constant)
f(x) xx d)f(
sec2 kx k1 tan kx
xtan xsecln
xcot xsinln
xcosec )tan(ln,cotcosecln 21 xxx +−
xsec )tan(ln,tansecln 41
21 π++ xxx
−= xxuvuvx
xvu d
ddd
dd
6 Edexcel AS/A level Mathematics Formulae List: Core Mathematics C3 – Issue 1 – September 2009
Core Mathematics C3
Candidates sitting C3 may also require those formulae listed under Core Mathematics C1 and C2.
Logarithms and exponentials
xax a=lne
Trigonometric identities
BABABA sincoscossin)(sin ±=±BABABA sinsincoscos)(cos =±
))(( tantan1tantan)(tan 2
1 π+≠±±=± kBABABABA
2cos
2sin2sinsin BABABA −+=+
2sin
2cos2sinsin BABABA −+=−
2cos
2cos2coscos BABABA −+=+
2sin
2sin2coscos BABABA −+−=−
Differentiation
f(x) f ′(x)
tan kx k sec2 kx
sec x sec x tan x
cot x –cosec2 x
cosec x –cosec x cot x
)g()f(
xx
))(g(
)(g)f( )g()(f2x
xxxx ′−′
Edexcel AS/A level Mathematics Formulae List: Core Mathematics C2 – Issue 1 – September 2009 5
Core Mathematics C2
Candidates sitting C2 may also require those formulae listed under Core Mathematics C1.
Cosine rule
a2 = b2 + c2 – 2bc cos A
Binomial series
21
)( 221 nrrnnnnn bbarn
ban
ban
aba ++++++=+ −−− (n ∈ )
where)!(!
!C rnr
nrn
rn
−==
∈<+×××
+−−++×−++=+ nxx
rrnnnxnnnxx rn ,1(
21)1()1(
21)1(1)1( 2 )
Logarithms and exponentials
ax
xb
ba log
loglog =
Geometric series
un = arn − 1
Sn = r ra n
−−
1)1(
S∞ = r
a−1
for ⏐r⏐ < 1
Numerical integration
The trapezium rule: b
a
xy d ≈ 21 h{(y0 + yn) + 2(y1 + y2 + ... + yn – 1)}, where
nabh −=
4 Edexcel AS/A level Mathematics Formulae List: Core Mathematics C1 – Issue 1 – September 2009
Core Mathematics C1
Mensuration
Surface area of sphere = 4π r 2
Area of curved surface of cone = π r × slant height
Arithmetic series
un = a + (n – 1)d
Sn = 21 n(a + l) =
21 n[2a + (n − 1)d]
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