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A-Level · Pure Mathematics

P8 Pure: Integration

EDEXCEL 9MA0 · Calculator allowed · about 140 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question tests integration of expressions with fractional and negative indices.

(a)Find integral of (3x² - 4x^-1/2 + 5) dx, giving your answer in simplified form.(3)

(b)Show that integral of (2x - 3)/√x dx = (4/3)x^3/2 - 6x^1/2 + c(3)

(c)Hence evaluate the definite integral, from x = 1 to x = 4, of (2x - 3)/√x dx, giving your answer as an exact value.(2)

(Total for Question 1 is 8 marks)

This question tests integration of e^kx and 1/x, and finding a particular solution using a given point.

(a)Find integral of (e^2x + 3/x) dx, for x > 0.(2)

(b)A curve C has dy/dx = 6x² - e^-x, and C passes through the point (0, 5). Find y in terms of x.(4)

(c)Hence find the gradient of C at the point where x = 0.(2)

(Total for Question 2 is 8 marks)

This question tests integration by inspection (reverse chain rule).

(a)Find integral of (2x + 1)⁵ dx.(3)

(b)Find integral of x/(x² + 4) dx.(3)

(c)Find integral of cos(3x - 1) dx.(3)

(Total for Question 3 is 9 marks)

The curve C has equation y = x³ - 4x.

(a)Show that the definite integral, from x = 0 to x = 2, of (x³ - 4x) dx equals -4.(2)

(b)Explain why the value found in part (a) is not the area of the region bounded by C and the x-axis between x = 0 and x = 2, and state the area of this region.(2)

(c)Find the total area enclosed between the curve C and the x-axis for 0 ≤ x ≤ 3.(5)

(Total for Question 4 is 9 marks)

The curve C has equation y = 6x - x², and the line l has equation y = 2x.

(a)Find the coordinates of the points of intersection of C and l.(3)

(b)Find the area of the region enclosed by C and l.(4)

(Total for Question 5 is 7 marks)

The table shows values of y = √1 + x³, for 0 ≤ x ≤ 2.
x : 0, 0.5, 1, 1.5, 2
y : 1, 1.0607, ?, 2.0917, 3

(a)Complete the table by finding the value of y at x = 1, giving your answer to 4 decimal places.(2)

(b)Using the trapezium rule with all five values in the table (strip width h = 0.5), find an estimate for the definite integral, from x = 0 to x = 2, of √1 + x³ dx, giving your answer to 3 significant figures.(3)

(c)State, with a reason, whether the trapezium rule estimate found in part (b) is an overestimate or an underestimate of the true value of the integral.(2)

(Total for Question 6 is 7 marks)

This question tests integration by substitution.

(a)Using the substitution u = 2x - 1, find integral of x*√2x - 1 dx.(5)

(b)Hence evaluate the definite integral, from x = 1 to x = 5, of x*√2x - 1 dx, giving your answer as an exact fraction.(3)

(Total for Question 7 is 8 marks)

This question tests integration of trigonometric functions using double angle identities.

(a)Using the identity cos(2A) = 1 - 2sin²(A), or otherwise, find integral of sin²(x) dx.(3)

(b)Find integral of cos²(3x) dx.(4)

(c)Hence evaluate the definite integral, from x = 0 to x = π/6, of cos²(3x) dx, giving your answer in terms of π.(3)

(Total for Question 8 is 10 marks)

This question tests integration by parts.

(a)Find integral of x*sin(2x) dx, using integration by parts.(5)

(b)Find integral of x² * ln(x) dx, for x > 0, using integration by parts.(5)

(Total for Question 9 is 10 marks)

f(x) = (5x - 14) / ((x - 1)(x - 4)), for x > 4.

(a)Express f(x) in the form A/(x - 1) + B/(x - 4), where A and B are integers to be found.(4)

(b)Hence find integral of f(x) dx.(4)

(c)Hence show that the definite integral, from x = 5 to x = 6, of f(x) dx equals ln(k), and find the exact value of k.(4)

(Total for Question 10 is 12 marks)

The region R is bounded by the curve y = e^x, the x-axis, and the lines x = 0 and x = 1.

(a)Find the volume of the solid generated when R is rotated 2*π radians about the x-axis, giving your answer as an exact multiple of π.(5)

(b)Give your answer to part (a) as a decimal, correct to 3 significant figures.(3)

(Total for Question 11 is 8 marks)

The temperature, θ degrees Celsius, of a cup of tea t minutes after it is made satisfies the differential equation dtheta/dt = -k(θ - 18), where k is a positive constant and 18 is the room temperature. When t = 0, θ = 90.

(a)Using separation of variables, show that θ = 18 + 72*e^-kt.(5)

(b)Given that θ = 60 when t = 5, show that k = (1/5) ln(12/7).(3)

(c)Find the time taken, in minutes, for the temperature of the tea to fall to 25 degrees Celsius, giving your answer to 3 significant figures.(3)

(Total for Question 12 is 11 marks)

A curve C has parametric equations x = t², y = 3 - t, for 0 ≤ t ≤ 3.

(a)Find dx/dt.(1)

(b)Show that C crosses the x-axis at the point where t = 3, and find the coordinates of this point.(2)

(c)Use the formula Area = integral of y*(dx/dt) dt to find the area of the region bounded by the curve C, the x-axis and the y-axis.(5)

(d)Verify your answer to part (c) by finding the Cartesian equation of C in the form y = f(x), and evaluating the appropriate definite integral in x.(4)

(Total for Question 13 is 12 marks)

Mark scheme · P8 Pure: Integration

Question 1

(a) M1 attempt to integrate at least two terms, index raised by 1 throughout
(a) A1 x³ - 8x^1/2 with both coefficients correct oe
(a) B1 +5x + c present and correct
(a) Answer: x³ - 8sqrt(x) + 5x + c
(b) M1 rewrite integrand as 2x^1/2 - 3x^-1/2
(b) M1 integrate to (4/3)x^3/2 and -6x^1/2, indices raised by 1 (dep)
(b) A1 cso, fully correct with +c matching the printed result
(b) Answer: (4/3)x^3/2 - 6x^1/2 + c (given, cso)
(c) M1 substitute limits x = 4 and x = 1 into the result from part (b)
(c) A1 10/3 oe, cao
(c) Answer: 10/3

Question 2

(a) M1 (1/2)e^2x term correct
(a) A1 3 ln x + c, all correct oe
(a) Answer: (1/2)e^2x + 3 ln x + c
(b) M1 integrate 6x² to 2x³
(b) M1 integrate -e^-x to e^-x, sign correct
(b) M1 substitute x = 0, y = 5 to find the constant
(b) A1 y = 2x³ + e^-x + 4, cao
(b) Answer: y = 2x³ + e^-x + 4
(c) M1 substitute x = 0 into 6x² - e^-x
(c) A1 -1, cao
(c) Answer: -1

Question 3

(a) M1 attempt at (2x+1)⁶ form
(a) M1 divide by 12 (their coefficient from the chain rule), dep
(a) A1 (2x+1)⁶ / 12 + c, cao
(a) Answer: (2x + 1)⁶ / 12 + c
(b) M1 recognise numerator is (1/2) x derivative of denominator
(b) A1 (1/2) ln(x² + 4)
(b) B1 + c present
(b) Answer: (1/2) ln(x² + 4) + c
(c) M1 sin(3x - 1) form seen
(c) A1 coefficient 1/3 correct
(c) B1 + c present
(c) Answer: (1/3) sin(3x - 1) + c

Question 4

(a) M1 integrate to x⁴/4 - 2x²
(a) A1 cso, substitute limits correctly to obtain -4
(a) Answer: -4 (given, cso)
(b) B1 correct explanation: y = x³ - 4x is negative (curve below the x-axis) for 0 < x < 2, so the integral is negative
(b) B1 area = 4 (square units)
(b) Answer: Area = 4 square units
(c) M1 split the region at x = 2, using the root of x³ - 4x = 0
(c) M1 find the definite integral from x = 2 to x = 3
(c) A1 value 6.25 (or 25/4) correct
(c) M1 combine |-4| + 6.25, taking the absolute value of the negative part
(c) A1 total area = 41/4 (or 10.25), cao
(c) Answer: 41/4 square units (10.25)

Question 5

(a) M1 set 6x - x² = 2x and rearrange to form a quadratic in x
(a) M1 solve by factorising to obtain x = 0 and x = 4
(a) A1 both coordinates (0, 0) and (4, 8) correct
(a) Answer: (0, 0) and (4, 8)
(b) M1 form the difference (curve minus line) = 4x - x²
(b) M1 integrate to 2x² - x³/3
(b) M1 correct use of limits x = 0 and x = 4
(b) A1 32/3 (or awrt 10.7), cao
(b) Answer: 32/3 square units (awrt 10.7)

Question 6

(a) M1 substitute x = 1 into √1 + x³
(a) A1 awrt 1.4142
(a) Answer: 1.4142 (awrt)
(b) M1 correct trapezium rule structure: (h/2)[(y0 + y4) + 2(y1 + y2 + y3)]
(b) M1 correct substitution of h = 0.5 and all five table values
(b) A1 awrt 3.28
(b) Answer: 3.28 (awrt)
(c) B1 overestimate
(c) B1 correct reason: the curve is convex (concave up, gradient increasing) on this interval, so the trapezia lie above the curve
(c) Answer: Overestimate, since the curve is convex (concave up) over [0, 2], so each trapezium lies above the curve.

Question 7

(a) M1 let u = 2x - 1, express x = (u+1)/2 and dx = du/2
(a) M1 substitute to obtain (1/4) integral of (u^3/2 + u^1/2) du
(a) M1 integrate to (1/4)[(2/5)u^5/2 + (2/3)u^3/2]
(a) A1 simplify to (1/10)u^5/2 + (1/6)u^3/2 (+c)
(a) A1 substitute back u = 2x-1, cao
(a) Answer: (1/10)(2x - 1)^5/2 + (1/6)(2x - 1)^3/2 + c
(b) M1 substitute x = 5 (u = 9) into their result from part (a)
(b) M1 substitute x = 1 (u = 1) and subtract
(b) A1 428/15 oe (awrt 28.5), cao
(b) Answer: 428/15 (awrt 28.5)

Question 8

(a) M1 rewrite sin²(x) as (1 - cos2x)/2
(a) M1 integrate term by term
(a) A1 (1/2)x - (1/4) sin(2x) + c, cao
(a) Answer: (1/2)x - (1/4) sin(2x) + c
(b) M1 use identity cos²(3x) = (1 + cos6x)/2
(b) M1 integrate the constant term to (1/2)x
(b) M1 integrate the cos6x term, dividing by 6 (chain rule factor)
(b) A1 (1/2)x + (1/12) sin(6x) + c, cao
(b) Answer: (1/2)x + (1/12) sin(6x) + c
(c) M1 substitute upper limit x = π/6, using sin(π) = 0
(c) M1 substitute lower limit x = 0
(c) A1 π/12, cao
(c) Answer: π/12

Question 9

(a) M1 choose u = x, dv/dx = sin(2x), correctly giving du/dx = 1 and v = -(1/2)cos(2x)
(a) M1 apply the by-parts formula uv - integral of v du
(a) A1 correct expression -x/2 cos(2x) + (1/2) integral of cos(2x) dx
(a) A1 integrate the remaining term to (1/4) sin(2x)
(a) A1 -(x/2) cos(2x) + (1/4) sin(2x) + c, cao
(a) Answer: -(x/2) cos(2x) + (1/4) sin(2x) + c
(b) M1 choose u = ln x, dv/dx = x², correctly giving du/dx = 1/x and v = x³/3
(b) M1 apply the by-parts formula uv - integral of v du
(b) A1 correct expression (x³/3) ln x - (1/3) integral of x² dx
(b) M1 integrate the remaining integral to (1/9)x³
(b) A1 (x³/3) ln x - (1/9)x³ + c, cao
(b) Answer: (x³/3) ln(x) - (1/9)x³ + c

Question 10

(a) M1 set up 5x - 14 = A(x - 4) + B(x - 1)
(a) M1 substitute x = 4 or x = 1 to find one constant
(a) A1 one of A = 3, B = 2 correct
(a) A1 both A = 3 and B = 2 correct
(a) Answer: A = 3, B = 2
(b) M1 integrate A/(x-1) term to A ln(x-1)
(b) M1 integrate B/(x-4) term to B ln(x-4)
(b) A1 3 ln(x - 1) + 2 ln(x - 4) all correct
(b) B1 + c present
(b) Answer: 3 ln(x - 1) + 2 ln(x - 4) + c
(c) M1 substitute x = 6 and x = 5 into their antiderivative from part (b)
(c) M1 combine logs using the laws of logarithms: 3ln5 + 2ln2 - 3ln4
(c) A1 simplify to ln(125/16), oe
(c) A1 k = 125/16, cao
(c) Answer: k = 125/16

Question 11

(a) M1 V = π * integral of y² dx, with y² = e^2x
(a) M1 integrate to (1/2)e^2x
(a) M1 substitute upper limit x = 1
(a) A1 subtract the value at x = 0 correctly
(a) A1 (π/2)(e² - 1), oe, cao
(a) Answer: (π/2)(e² - 1)
(b) M1 correct substitution of e² = 7.389... into their expression
(b) A1 awrt 10.0
(b) Answer: 10.0 (awrt)

Question 12

(a) M1 separate variables correctly: integral of 1/(θ-18) dtheta = integral of -k dt
(a) A1 ln(θ - 18) = -kt (+ constant)
(a) M1 rearrange (exponentiate) to θ - 18 = A*e^-kt
(a) M1 use t = 0, θ = 90 to find A = 72
(a) A1 cso, θ = 18 + 72*e^-kt
(a) Answer: θ = 18 + 72*e^-kt (given, cso)
(b) M1 substitute t = 5, θ = 60 and rearrange to e^-5k = 7/12
(b) M1 take ln of both sides
(b) A1 cso, k = (1/5) ln(12/7)
(b) Answer: k = (1/5) ln(12/7) (given, cso)
(c) M1 substitute θ = 25 into θ = 18+72e^-kt and rearrange to e^-kt = 7/72
(c) M1 take logs and use their value of k (ft) to solve for t
(c) A1 awrt 21.6 (minutes)
(c) Answer: t = awrt 21.6 minutes

Question 13

(a) B1 dx/dt = 2t
(a) Answer: dx/dt = 2t
(b) B1 cso, y = 3 - 3 = 0 when t = 3
(b) B1 coordinates (9, 0)
(b) Answer: (9, 0)
(c) M1 Area = integral, from t=0 to t=3, of (3-t)(2t) dt, correct limits
(c) M1 expand integrand to 6t - 2t²
(c) M1 integrate to 3t² - (2/3)t³
(c) M1 substitute limits t = 3 and t = 0
(c) A1 area = 9, cao
(c) Answer: 9 square units
(d) M1 find Cartesian equation y = 3 - √x, using t = √x
(d) M1 set up integral, from x = 0 to x = 9, of (3 - √x) dx
(d) M1 integrate to 3x - (2/3)x^3/2
(d) A1 evaluate to 9, confirming part (c)
(d) Answer: Area = 9 (confirmed)

Pure: Integration - Worksheets, Questions and Revision