Evaluate the definite integral, from x = 1 to x = 3, of 2x dx.
(Total for Question 5 is 1 mark)
6
Find the integral of √x with respect to x.
(Total for Question 6 is 1 mark)
7
Find the integral of cos(2x) with respect to x.
(Total for Question 7 is 1 mark)
8
Find the integral of (6x2 - 4x-1/2) with respect to x.
(Total for Question 8 is 2 marks)
9
Evaluate the definite integral, from x = 0 to x = 2, of (3x2 + 1) dx.
(Total for Question 9 is 2 marks)
10
Find the integral of (3/x + 2ex) with respect to x, for x > 0.
(Total for Question 10 is 2 marks)
11
Using inspection (the reverse chain rule), find the integral of (2x + 5)3 with respect to x.
(Total for Question 11 is 2 marks)
12
Using inspection (the reverse chain rule), find the integral of x/(x2 + 9) with respect to x, giving your answer in terms of a natural logarithm.
(Total for Question 12 is 2 marks)
13
Find the integral of sin(4x - 1) with respect to x.
(Total for Question 13 is 2 marks)
14
The curve C has equation y = 8x - x2, and the line l has equation y = 3x. Find the coordinates of the points of intersection of C and l, and find the area of the region enclosed by C and l.
(Total for Question 14 is 3 marks)
15
Find the total area enclosed between the curve C: y = x2 - 4x and the x-axis, for 0 ≤ x ≤ 6.
(Total for Question 15 is 3 marks)
16
Using the substitution u = 2x + 1, find the integral of x*√2x + 1 with respect to x. Hence evaluate the definite integral, from x = 0 to x = 4, of x*√2x + 1 dx, giving your answer as an exact value.
(Total for Question 16 is 3 marks)
17
Find the integral of x*cos(3x) with respect to x, using integration by parts. Hence evaluate the definite integral, from x = 0 to x = π/6, of x*cos(3x) dx, giving your answer as an exact value in terms of π.
(Total for Question 17 is 4 marks)
18
f(x) = (7x - 16) / ((x - 2)(x - 3)), for x > 3. Express f(x) in the form A/(x - 2) + B/(x - 3), where A and B are integers to be found. Hence find the integral of f(x) with respect to x, and show that the definite integral, from x = 4 to x = 5, of f(x) dx equals ln(k), finding the exact value of k.
(Total for Question 18 is 4 marks)
19
The region R is bounded by the curve y = √2x + 1, the x-axis, and the lines x = 0 and x = 4. 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. Give your answer also as a decimal, correct to 3 significant figures.
(Total for Question 19 is 4 marks)
Mark scheme · P8D Pure: Integration: Fluency and Exam Drill
Question 1
B1 (1/4)x4 + c cao
Answer: (1/4)x4 + c
Question 2
B1 2x2 - 3x + c cao
Answer: 2x2 - 3x + c
Question 3
B1 -x-1 + c cao, oe -1/x + c
Answer: -1/x + c
Question 4
B1 (1/4)e4x + c cao
Answer: (1/4)e4x + c
Question 5
B1 8 cao
Answer: 8
Question 6
B1 (2/3)x3/2 + c cao
Answer: (2/3)x3/2 + c
Question 7
B1 (1/2)sin(2x) + c cao
Answer: (1/2)sin(2x) + c
Question 8
M1 integrates each term, raising the power by 1 and dividing by the new power
A1 2x3 - 8x1/2 + c cao, oe 2x3 - 8sqrt(x) + c
Answer: 2x3 - 8sqrt(x) + c
Question 9
M1 finds the antiderivative x3 + x
A1 10 cao
Answer: 10
Question 10
M1 integrates 3/x to 3ln(x)
A1 3ln(x) + 2ex + c cao
Answer: 3ln(x) + 2ex + c
Question 11
M1 recognises the form k(2x+5)4 and finds k by differentiating to check
A1 (1/8)(2x+5)4 + c cao
Answer: (1/8)(2x+5)4 + c
Question 12
M1 recognises that the numerator is (1/2) of the derivative of the denominator
A1 (1/2)ln(x2+9) + c cao
Answer: (1/2)ln(x2+9) + c
Question 13
M1 recognises the form -k*cos(4x-1) and finds k using the chain rule
A1 -(1/4)cos(4x-1) + c cao
Answer: -(1/4)cos(4x-1) + c
Question 14
M1 sets 8x - x2 = 3x and solves to find x = 0 and x = 5
M1 sets up and integrates the area as the integral of (5x - x2) dx
A1 125/6 cao, oe awrt 20.8
Answer: Points (0,0) and (5,15); area = 125/6
Question 15
M1 finds the roots x = 0 and x = 4, and recognises the curve is below the axis for 0 < x < 4
M1 integrates the two regions separately, from 0 to 4 and from 4 to 6, taking the modulus of the first
A1 64/3 cao, oe awrt 21.3
Answer: Area = 64/3
Question 16
M1 substitutes x = (u-1)/2, dx = (1/2)du to form (1/4) times the integral of (u-1)*u1/2 du
M1 integrates to (1/10)u5/2 - (1/6)u3/2 + c and substitutes the limits u=1 (x=0) and u=9 (x=4)
A1 298/15 cao, oe awrt 19.9
Answer: 298/15
Question 17
M1 sets u = x, dv/dx = cos(3x), so du/dx = 1, v = (1/3)sin(3x), and applies the by-parts formula
A1 (x/3)sin(3x) + (1/9)cos(3x) + c cao
M1 substitutes the limits x = π/6 and x = 0
A1 (π-2)/18 cao, oe awrt 0.0634
Answer: (π-2)/18
Question 18
M1 forms 7x - 16 = A(x-3) + B(x-2) and substitutes x=2 and x=3 (or compares coefficients)
A1 A = 2, B = 5 cao
M1 integrates to 2ln(x-2) + 5ln(x-3) + c and substitutes the limits 4 and 5
A1 k = 72 cao
Answer: A = 2, B = 5; k = 72
Question 19
M1 sets up the volume as π times the integral of (2x+1) dx from 0 to 4
M1 integrates to π*[x2 + x] and substitutes the limits 0 and 4