Pure: Integration Depth - Worksheets, Questions and Revision

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A-Level · Pure Mathematics

P15 Pure: Integration Depth

EDEXCEL 9MA0 · Calculator allowed · about 140 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Find integral of 6x(x² + 5)² dx.

(Total for Question 1 is 3 marks)

Using the substitution u = 4 + x², find the exact value of the definite integral, from x = 0 to x = √5, of x/√4 + x² dx, showing each step of your working.

(Total for Question 2 is 5 marks)

A curve satisfies the differential equation dy/dx = (2x + 1)/y², where y > 0. The curve passes through the point (0, 2). Find y in terms of x, giving your answer in the form y = (f(x))^1/3.

(Total for Question 3 is 5 marks)

Find integral of (sin(x))³ * (cos(x))² dx.

(Total for Question 4 is 5 marks)

f(x) = x * e^2x.

(a)Find integral of f(x) dx.(4)

(b)Hence find the exact value of the definite integral, from x = 0 to x = 1, of f(x) dx.(3)

(c)Hence, or otherwise, write down the value of the definite integral, from x = 0 to x = 1, of f(x) dx, correct to 3 significant figures.(1)

(Total for Question 5 is 8 marks)

Find the exact value of the definite integral, from x = 1 to x = e, of x² * ln(x) dx, giving your answer in the form (1/9)*(a*e³ + b), where a and b are integers to be found.

(Total for Question 6 is 6 marks)

The curve C has equation y = 1/√9 - x², for -3 < x < 3.

(a)Find integral of 1/√9 - x² dx.(2)

(b)Hence show that the definite integral, from x = 0 to x = 3/2, of 1/√9 - x² dx, equals π/6.(3)

(Total for Question 7 is 5 marks)

f(x) = (8x + 7) / ((x - 1)(x + 4)).

(a)Express f(x) in the form A/(x - 1) + B/(x + 4), where A and B are constants to be found.(3)

(b)Hence find integral of f(x) dx.(3)

(c)Find the exact value of the definite integral, from x = 2 to x = 6, of f(x) dx, giving your answer in the form p*ln(5) + q*ln(3), where p and q are integers to be found.(4)

(Total for Question 8 is 10 marks)

g(x) = (x² + 5x - 5) / ((x - 2)² * (x + 1)).

(a)Express g(x) in the form A/(x - 2) + B/(x - 2)² + C/(x + 1), where A, B and C are constants to be found.(4)

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

(c)Find the exact value of the definite integral, from x = 3 to x = 4, of g(x) dx, giving your answer in the form a + ln(b), where a is a rational number and b is a fraction in lowest terms.(3)

(Total for Question 9 is 11 marks)

The curve C has equation y = 4x² - x³. The line l has equation y = 3x.

(a)Show that C and l intersect at x = 0, x = 1 and x = 3.(3)

(b)Find the area of the finite region bounded by C and l for 1 ≤ x ≤ 3.(6)

(c)Find the area of the finite region bounded by C and l for 0 ≤ x ≤ 1.(4)

(Total for Question 10 is 13 marks)

The curve C has equation y = x/(1 + x²), for x ≥ 0.

(a)Show that integral of x/(1 + x²) dx = (1/2)*ln(1 + x²) + c.(2)

(b)Hence find the exact area of the region bounded by C, the x-axis and the lines x = 0 and x = √3, giving your answer in the form k*ln(2), where k is a constant to be found.(3)

(c)Find the exact area of the region bounded by C, the x-axis and the lines x = √3 and x = 3, giving your answer in the form (1/2)*ln(p/q), where p and q are integers in lowest terms.(3)

(Total for Question 11 is 8 marks)

Find integral of e^2x*sin(x) dx. You may use integration by parts twice.

(Total for Question 12 is 6 marks)

The number of members, P, of a newly-launched online chess club is modelled by the differential equation dP/dt = P(80 - P)/200, where t is the time in weeks after the club's launch and 0 < P < 80. When t = 0, P = 10.

(a)Show that 1/(P(80 - P)) = (1/80)*(1/P + 1/(80 - P)).(2)

(b)By separating the variables and using the result from part (a), show that the general solution of the differential equation can be written as ln(P/(80 - P)) = 0.4*t + c, where c is an arbitrary constant.(4)

(c)Given that P = 10 when t = 0, find the value of c, and hence show that P = 80/(1 + 7*e^-0.4*t).(4)

(d)Find the time, in weeks, to 3 significant figures, for the club's membership to reach 40.(3)

(e)State one limitation of this model for predicting the club's membership over a long period of time.(1)

(Total for Question 13 is 14 marks)

Mark scheme · P15 Pure: Integration Depth

Question 1

M1 recognise the integrand as a multiple of the derivative of (x²+5), i.e. attempt the substitution u=x²+5 or reverse chain rule
M1 integrate to the form k(x²+5)³
A1 (x²+5)³ + c cao
Answer: (x² + 5)³ + c

Question 2

M1 du/dx = 2x, so x dx = (1/2) du (oe)
M1 rewrite the integral in terms of u as (1/2) * integral of u^-1/2 du
M1 integrate to obtain √u, i.e. √4+x²
A1 correctly change/back-substitute the limits: u=9 when x=√5, u=4 when x=0 (or equivalent evaluation in x)
A1 1 cso
Answer: 1

Question 3

M1 separate the variables: y² dy = (2x+1) dx (oe, condone missing differentials)
A1 integrate correctly: y³/3 = x² + x (+c)
M1 substitute x=0, y=2 to find the constant, c = 8/3
A1 y³ = 3x² + 3x + 8 oe (e.g. multiplying through by 3)
A1 y = (3x² + 3x + 8)^1/3 cao
Answer: y = (3x² + 3x + 8)^1/3

Question 4

M1 use (sin(x))² = 1 - (cos(x))² to rewrite the integrand as sin(x)*((cos(x))² - (cos(x))⁴) oe
M1 substitute u = cos(x), du = -sin(x) dx (oe)
M1 integrate to the form u⁵/5 - u³/3 (or equivalent sign-consistent form)
A1 correct unsimplified answer in terms of u or x
A1 (cos(x))⁵/5 - (cos(x))³/3 + c cao
Answer: (cos(x))⁵/5 - (cos(x))³/3 + c

Question 5

(a) M1 attempt integration by parts with u=x, dv/dx=e^2x
(a) A1 v = (1/2)e^2x identified (du=dx may be implied)
(a) M1 correct application: (x/2)*e^2x - integral of (1/2)*e^2x dx
(a) A1 (x/2)*e^2x - (1/4)*e^2x (+c) oe cao
(a) Answer: (x/2)*e^2x - (1/4)*e^2x + c
(b) M1 substitute x=1 into their answer from part (a) (ft)
(b) M1 substitute x=0 into their answer from part (a) to obtain -1/4 (ft)
(b) A1 (e² + 1)/4 cao
(b) Answer: (e² + 1)/4
(c) B1 awrt 2.10 (ft from part (b))
(c) Answer: 2.10 (3 s.f.)

Question 6

M1 attempt integration by parts with u=ln(x), dv/dx=x²
A1 v = x³/3 identified (du=(1/x)dx may be implied)
M1 correct application: (x³/3)*ln(x) - integral of (x³/3)*(1/x) dx
A1 (x³/3)*ln(x) - x³/9 (+c) oe
M1 substitute limits x=e and x=1, using ln(e)=1 and ln(1)=0
A1 (1/9)*(2*e³ + 1), i.e. a=2 and b=1, cao
Answer: (2e³ + 1)/9, i.e. a = 2, b = 1

Question 7

(a) B1 recognise the standard result leading to an arcsin(x/a) form, with a=3 identified
(a) A1 arcsin(x/3) + c cao
(a) Answer: arcsin(x/3) + c
(b) M1 substitute x=3/2 and x=0 into their answer from part (a) (ft)
(b) A1 arcsin(1/2) - arcsin(0) obtained
(b) A1 = π/6 cso
(b) Answer: π/6

Question 8

(a) M1 form the identity 8x+7 = A(x+4) + B(x-1) (or use the cover-up method directly)
(a) A1 A = 3
(a) A1 B = 5
(a) Answer: f(x) = 3/(x - 1) + 5/(x + 4)
(b) M1 integrate each partial fraction term from (a) to a ln form (ft their A, B)
(b) A1 3*ln|x-1| (ft)
(b) A1 3*ln|x-1| + 5*ln|x+4| (+c) cao
(b) Answer: 3*ln|x - 1| + 5*ln|x + 4| + c
(c) M1 substitute limits x=6 and x=2 into their answer from (b) (ft)
(c) M1 use the laws of logarithms to combine terms, e.g. writing ln(10) and ln(6) in terms of ln(2), ln(3) and ln(5)
(c) A1 correct simplification to 8*ln(5) - 5*ln(3) (unsimplified intermediate working may be seen)
(c) A1 p = 8, q = -5 cao
(c) Answer: 8*ln(5) - 5*ln(3), i.e. p = 8, q = -5

Question 9

(a) M1 form the identity x²+5x-5 = A(x-2)(x+1) + B(x+1) + C(x-2)² (or use the cover-up method for B and C)
(a) A1 B = 3 (from substituting x=2)
(a) A1 C = -1 (from substituting x=-1)
(a) A1 A = 2 (e.g. by comparing coefficients of x²), all three values stated
(a) Answer: g(x) = 2/(x - 2) + 3/(x - 2)² - 1/(x + 1)
(b) M1 integrate the A/(x-2) term to A*ln|x-2| (ft their A)
(b) M1 integrate the B/(x-2)² term to -B/(x-2) (ft their B)
(b) A1 integrate the C/(x+1) term to C*ln|x+1| (ft their C)
(b) A1 2*ln|x-2| - 3/(x-2) - ln|x+1| (+c) cao
(b) Answer: 2*ln|x - 2| - 3/(x - 2) - ln|x + 1| + c
(c) M1 substitute limits x=4 and x=3 into their answer from (b) (ft)
(c) M1 use the laws of logarithms to combine the ln terms into a single logarithm
(c) A1 3/2 + ln(16/5) cao (equivalently 3/2 + 4ln(2) - ln(5))
(c) Answer: 3/2 + ln(16/5)

Question 10

(a) M1 set 4x²-x³=3x and rearrange to x³-4x²+3x=0 (oe)
(a) M1 factorise to x(x-1)(x-3)=0 (or equivalent correct factorisation)
(a) A1 x = 0, x = 1 and x = 3 all stated cso
(a) Answer: x = 0, x = 1, x = 3
(b) M1 set up the definite integral, from x = 1 to x = 3, of (4x²-x³-3x) dx
(b) M1 integrate to (4/3)x³ - (1/4)x⁴ - (3/2)x²
(b) A1 correct antiderivative, all three terms correct
(b) M1 substitute limits x=3 and x=1 into their antiderivative
(b) A1 correct evaluation: 9/4 at x=3 and -5/12 at x=1 (or equivalent decimals), seen
(b) A1 8/3 cao
(b) Answer: 8/3
(c) M1 recognise that l is above C on this interval and set up the definite integral, from x = 0 to x = 1, of (3x-4x²+x³) dx (oe, condone use of |curve-line| reasoning)
(c) M1 integrate to (3/2)x² - (4/3)x³ + (1/4)x⁴
(c) A1 correct antiderivative, all three terms correct
(c) A1 5/12 cao
(c) Answer: 5/12

Question 11

(a) M1 recognise the integrand as (1/2) times the derivative of ln(1+x²) (or use the substitution u=1+x², du=2x dx)
(a) A1 (1/2)*ln(1+x²) + c cso
(a) Answer: (1/2)*ln(1 + x²) + c
(b) M1 substitute limits x=√3 and x=0 into the result from part (a) (ft)
(b) A1 (1/2)*ln(4) - (1/2)*ln(1) obtained
(b) A1 k = 1, i.e. ln(2), cao
(b) Answer: ln(2), i.e. k = 1
(c) M1 substitute limits x=3 and x=√3 into the result from part (a) (ft)
(c) A1 (1/2)*ln(10) - (1/2)*ln(4) obtained
(c) A1 (1/2)*ln(5/2), i.e. p=5, q=2, cao
(c) Answer: (1/2)*ln(5/2), i.e. p = 5, q = 2

Question 12

M1 let I = integral of e^2x*sin(x) dx and apply integration by parts once, e.g. u=sin(x), dv/dx=e^2x, to reach I=(1/2)e^2xsin(x) - (1/2)*integral of e^2x*cos(x) dx
M1 apply integration by parts a second time to the integral of e^2x*cos(x) dx, e.g. u=cos(x), dv/dx=e^2x, to reach (1/2)e^2xcos(x) + (1/2)*integral of e^2x*sin(x) dx
dM1 substitute the second result back to form an equation in I, e.g. I = (1/2)e^2xsin(x) - (1/4)e^2xcos(x) - (1/4)I, dependent on both previous method marks
M1 collect the I terms and solve for I, e.g. (5/4)I = (1/2)e^2xsin(x) - (1/4)e^2xcos(x)
A1 correct unsimplified form I = (2/5)e^2xsin(x) - (1/5)e^2xcos(x) (+c)
A1 I = (1/5)*e^2x*(2*sin(x) - cos(x)) + c, fully factorised, cao
Answer: (1/5)*e^2x*(2*sin(x) - cos(x)) + c

Question 13

(a) M1 combine the right-hand side over the common denominator 80*P*(80-P), reaching ((80-P)+P)/(80*P*(80-P))
(a) A1 simplify the numerator to 80, giving 1/(P(80-P)) cso
(a) Answer: 1/(P(80 - P)) = (1/80)*(1/P + 1/(80 - P))
(b) M1 separate the variables: integral of 1/(P(80-P)) dP = integral of (1/200) dt (oe)
(b) M1 use part (a) to integrate the left-hand side: (1/80)*(ln(P) - ln(80-P))
(b) A1 correct right-hand side integration, t/200 (+c)
(b) A1 rearrange (multiply both sides by 80) to ln(P/(80-P)) = 0.4*t + c (oe), cso
(b) Answer: ln(P/(80 - P)) = 0.4*t + c
(c) M1 substitute P=10, t=0 into the general solution to obtain c = ln(10/70) = ln(1/7), i.e. c = -ln(7)
(c) M1 rearrange to the exponential form P/(80-P) = e^{0.4t - ln(7)} = (1/7)*e^0.4t (oe)
(c) M1 rearrange algebraically to make P the subject, e.g. 7P = (80-P)*e^0.4t, leading to P*(e^0.4t+7) = 80*e^0.4t
(c) A1 divide through by e^0.4t to reach P = 80/(1 + 7*e^-0.4t) cso
(c) Answer: c = -ln(7); P = 80/(1 + 7*e^-0.4*t)
(d) M1 set P=40 in the formula from part (c) and rearrange to 7*e^-0.4t = 1 (oe)
(d) M1 take natural logs correctly to isolate t, e.g. t = ln(7)/0.4
(d) A1 awrt 4.86 (weeks) cao
(d) Answer: t = ln(7)/0.4 = 4.86 weeks (3 s.f.)
(e) B1 any valid limitation with brief justification, e.g. the model assumes membership approaches a fixed maximum of 80 and never exceeds it, but real membership could plateau at a different level, fluctuate seasonally, or exceed 80 if demand grows (accept any sensible, justified critique of the constant-capacity assumption)
(e) Answer: e.g. the model assumes a fixed maximum membership of 80 that can never be exceeded, which may not reflect real growth if the club expands its capacity or membership fluctuates due to factors the model does not capture

Question 1

3 marks

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Question 2

5 marks

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