Edexcel International A Level Maths · P4 · Pure Mathematics 4
Integration by substitution, by parts and with partial fractions
28 free, high-quality past-paper-style questions on Integration by substitution, by parts and with partial fractions for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 230 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P4-5-B-01P4 · A2Easy5 marks
The integral J is defined by J = ∫4x/(2x + 5)3 dx, x > −2.5.
(a)
Use the substitution u = 2x + 5 to show that J = ∫(u − 5)/u3 du.
(2)
(b)
Hence find J, giving your answer in terms of x.
(3)
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Answer
Mark
Notes
Part (a)
du/dx = 2, so dx = ½ du, and x = (u − 5)/2
B1
4x = 2(u − 5), so J = ∫2(u − 5)/u3 × ½ du = ∫(u − 5)/u3 du (shown, with both replacements seen)
A1
Part (b)
splits into u−2 − 5u−3 and raises a power by one
M1
−u−1 + (5/2)u−2
A1
−1/(2x + 5) + 5/(2(2x + 5)2) + c, in terms of x with + c
A1
Common error: Forgetting the factor from dx = ½ du. Every dx has to be replaced, and the ½ is what turns 2(u − 5) into (u − 5).
Common error: Taking u = x3. Then you would need ∫ln x dx as the first step and the integral only gets worse. With a ln x in the product, let u = ln x.
A rational function with two distinct linear factors in its denominator is defined by f(x) = (7x − 2)/((x + 1)(2x − 1)) for x > ½. One of the factors has a coefficient of x other than 1.
(a)
Express f(x) in partial fractions.
(3)
(b)
Hence find ∫f(x) dx.
(3)
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Answer
Mark
Notes
Part (a)
7x − 2 ≡ A(2x − 1) + B(x + 1)
M1
one of A = 3, B = 1
A1
3/(x + 1) + 1/(2x − 1)
A1
Part (b)
integrates to logarithms of the linear factors
M1
3 ln(x + 1) ft
A1
3 ln(x + 1) + ½ ln(2x − 1) + c ft (the ½ is required)
A1
Common error: Writing ∫1/(2x − 1) dx = ln(2x − 1). Differentiate that and you get 2/(2x − 1): divide by the coefficient of x.
Figure 1 shows the curve with equation y = 2x cos x, 0 ≤ x ≤ π/2. The curve meets the x-axis at the origin O and at the point A. The shaded region R is bounded by the curve and the x-axis between O and A.
(a)
Write down the exact x-coordinate of A.
(1)
(b)
Use integration by parts to find ∫2x cos x dx.
(3)
(c)
Hence find the exact area of R.
(2)
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Answer
Mark
Notes
Part (a)
cos x = 0 with x > 0, so x = π/2
B1
Part (b)
differentiates 2x and integrates cos x
M1
2x sin x − ∫2 sin x dx
A1
2x sin x + 2 cos x + c
A1
Part (c)
[2x sin x + 2 cos x]0π/2 = (π + 0) − (0 + 2)
M1
π − 2
A1
Common error: Getting the sign wrong in the second step. −∫2 sin x dx = −(−2 cos x) = +2 cos x; differentiate your answer to check that it gives back 2x cos x.
Figure 1 shows part of the curve C with equation y = 6/((x + 1)(x + 4)), x ≥ 0. C meets the y-axis at A(0, 1.5). The shaded region R is bounded by C, the coordinate axes and the line x = 2.
(a)
Express 6/((x + 1)(x + 4)) in partial fractions.
(2)
(b)
Hence show that the area of R is ln 4.
(4)
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Answer
Mark
Notes
Part (a)
6 ≡ A(x + 4) + B(x + 1), with x = −1 and x = −4 (or equating coefficients)
M1
2/(x + 1) − 2/(x + 4)
A1
Part (b)
integrates both terms to logarithms
M1
2 ln(x + 1) − 2 ln(x + 4)
A1
(2 ln 3 − 2 ln 6) − (2 ln 1 − 2 ln 4)
M1
= 2 ln(3 × 4/6) = 2 ln 2 = ln 4
A1
Exam tip: Collect the logarithms at the end with the laws of logs: k ln a = ln ak and ln a − ln b = ln(a/b). Leaving 2 ln 3 − 2 ln 6 + 2 ln 4 does not show the printed answer.
Figure 1 shows part of the curve C with equation y = e2x/(ex + 1). C passes through the points A(0, ½) and B(ln 3, 9/4). The shaded region R is bounded by C, the coordinate axes and the line x = ln 3.
(a)
Use the substitution u = ex + 1 to show that the area of R is given by ∫24 (1 − 1/u) du.
(4)
(b)
Hence find the exact area of R.
(2)
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Answer
Mark
Notes
Part (a)
du/dx = ex, so dx = du/(u − 1)
B1
e2x = (ex)2 = (u − 1)2
B1
(u − 1)2/u × 1/(u − 1) = (u − 1)/u = 1 − 1/u
M1
limits: x = 0 gives u = 2 and x = ln 3 gives u = 4
B1
Part (b)
[u − ln u]24 = (4 − ln 4) − (2 − ln 2)
M1
2 − ln 2
A1
Common error: Changing the integrand but keeping the x-limits 0 and ln 3. Once everything is in terms of u, the limits must be u-values too: u = e0 + 1 = 2 and u = eln 3 + 1 = 4.
Figure 1 shows part of the curve C with equation y = ln x, x > 0. C meets the x-axis at A(1, 0) and passes through B(4, ln 4). The straight line segment AB is drawn. The shaded region S lies between C and the chord AB.
(a)
Use integration by parts to find ∫ln x dx.
(3)
(b)
Find the exact area of S, giving your answer in the form p ln 2 + q, where p and q are integers.
(3)
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Answer
Mark
Notes
Part (a)
writes ln x as 1 × ln x, differentiating ln x and integrating 1
M1
x ln x − ∫x × (1/x) dx
A1
x ln x − x + c
A1
Part (b)
area under C from 1 to 4: (4 ln 4 − 4) − (0 − 1) = 8 ln 2 − 3
M1
area under AB (a triangle): ½ × 3 × ln 4 = 3 ln 2
B1
S = 5 ln 2 − 3
A1
Exam tip: ln x on its own is integrated by parts by writing it as 1 × ln x. It is worth learning the result x ln x − x + c, but show the parts step when the question asks for it.
The denominator of g(x) = (3x2 + 6)/((x + 2)(x − 1)2), x > 1, contains a repeated linear factor, so three partial fractions are needed and one of them does not integrate to a logarithm.
(a)
Given that g(x) ≡ A/(x + 2) + B/(x − 1) + C/(x − 1)2, find the values of the constants A, B and C.
(4)
(b)
Hence find the exact value of ∫24 g(x) dx, giving your answer in the form p + ln q, where p is an integer and q is a rational number.
(5)
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Answer
Mark
Notes
Part (a)
3x2 + 6 ≡ A(x − 1)2 + B(x + 2)(x − 1) + C(x + 2)
M1
A = 2 (from x = −2)
A1
C = 3 (from x = 1)
A1
B = 1 (e.g. comparing x2: 3 = A + B)
A1
Part (b)
integrates the linear terms to logarithms: 2 ln(x + 2) + ln(x − 1)
M1
integrates the squared term to a power: −3/(x − 1)
The numerator and denominator of h(x) = (2x2 − 2x + 2)/((x − 1)(x + 1)), x > 1, are both quadratics, so h is an improper fraction and needs a constant term before its partial fractions.
(a)
Express h(x) in the form A + B/(x − 1) + C/(x + 1), where A, B and C are constants.
(4)
(b)
Hence find the exact value of ∫23 h(x) dx, giving your answer in the form a + ln b, where a is an integer and b is a rational number.
(4)
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Answer
Mark
Notes
Part (a)
recognises an improper fraction and finds A = 2 by division or comparing x2 terms
2 + ln 2 + 3 ln(3/4), using the laws of logarithms
M1
2 + ln(27/32)
A1
Exam tip: When the degree of the numerator is at least the degree of the denominator, the partial fractions need a polynomial part (here a constant) as well. Missing it leaves an identity that cannot hold.
Each of the following integrals needs a different method. In each case, find the integral.
(a)
∫cos x/(3 + sin x) dx
(2)
(b)
∫x sin 4x dx
(3)
(c)
∫(x + 5)/(x2 + x − 2) dx, x > 1
(3)
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Answer
Mark
Notes
Part (a)
recognises the numerator as the derivative of the denominator (f′/f), or uses u = 3 + sin x
M1
ln(3 + sin x) + c (no modulus needed since 3 + sin x > 0)
A1
Part (b)
parts with u = x, v = −¼ cos 4x
M1
−¼x cos 4x + ∫¼ cos 4x dx
A1
−¼x cos 4x + (1/16) sin 4x + c
A1
Part (c)
factorises x2 + x − 2 = (x − 1)(x + 2) and sets up partial fractions
M1
2/(x − 1) − 1/(x + 2)
A1
2 ln(x − 1) − ln(x + 2) + c ft
A1
Exam tip: Before choosing parts or partial fractions, check whether the numerator is (a multiple of) the derivative of the denominator. If it is, the answer is a logarithm in one line.
A particle P moves along a straight line. At time t seconds, t ≥ 0, the speed of P is v m s−1, where v = 12/((t + 1)(t + 3)). P does not change direction. Figure 1 shows the speed–time graph of P. The shaded region represents the distance travelled by P in the first 3 seconds.
(a)
Express 12/((t + 1)(t + 3)) in partial fractions.
(2)
(b)
Find the exact distance travelled by P in the first 3 seconds, giving your answer in the form k ln 2.
(3)
(c)
Show that the distance travelled by P in the first T seconds is 6 ln(3(T + 1)/(T + 3)) m. Hence find the exact value of T for which this distance is 3 ln 3 m.
(3 − √3)T = 3√3 − 3, so T = 3(√3 − 1)/(√3(√3 − 1)) = √3
A1
Exam tip: The distance is the area under a speed–time graph, so it is an integral of v. Because v stays positive in Figure 1, no part of the area has to be counted as negative.
Figure 1 shows the curve with equation y = sin 2xesin x, 0 ≤ x ≤ π/2. The curve meets the x-axis at the origin and at B(π/2, 0). The shaded region R is bounded by the curve and the x-axis.
(a)
Use the substitution u = sin x to show that the area of R is given by ∫01 2ueu du.
(4)
(b)
Hence, using integration by parts, show that the area of R is 2.
(4)
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Answer
Mark
Notes
Part (a)
du/dx = cos x, so dx = du/cos x
B1
sin 2x = 2 sin x cos x
B1
2 sin x cos xesin x × du/cos x = 2ueu du
M1
limits: x = 0 gives u = 0 and x = π/2 gives u = 1
B1
Part (b)
differentiates 2u and integrates eu
M1
2ueu − ∫2eu du = 2ueu − 2eu
A1
(2e − 2e) − (0 − 2)
M1
= 2
A1
Exam tip: When a substitution leaves a product such as ueu, a second method is needed. Recognising sin 2x as 2 sin x cos x is what makes the cos x cancel with dx = du/cos x.
Figure 1 shows the line with equation y = x and the curve C with equation y = x ln x, x > 0. C crosses the x-axis at A and meets the line at P. The shaded region R is bounded by the line from O to P, the curve from P to A, and the x-axis from A to O.
(a)
Find the exact coordinates of A and of P.
(3)
(b)
Use integration by parts to find ∫x ln x dx.
(3)
(c)
Show that the area of R is (e2 − 1)/4.
(3)
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Answer
Mark
Notes
Part (a)
x ln x = 0 with x > 0 gives ln x = 0, so A(1, 0)
B1
x ln x = x, so x(ln x − 1) = 0 with x ≠ 0
M1
ln x = 1, P(e, e)
A1
Part (b)
differentiates ln x and integrates x
M1
½x2 ln x − ∫½x2 × (1/x) dx
A1
½x2 ln x − ¼x2 + c
A1
Part (c)
area under the line from 0 to e: ½e2
B1
area under C from 1 to e: (½e2 − ¼e2) − (0 − ¼) = ¼e2 + ¼
M1
R = ½e2 − ¼e2 − ¼ = (e2 − 1)/4
A1
Common error: Choosing to integrate ln x and differentiate x. ln x has no simpler integral to start from, so it must be the part that is differentiated.
Figure 1 shows part of the curve C with equation y = 6/(x(x + 1)(x + 3)), x > 0. C passes through the point (1, 0.75). The shaded region R is bounded by C, the x-axis and the lines x = 1 and x = 3.
(a)
Express 6/(x(x + 1)(x + 3)) in partial fractions.
(3)
(b)
Hence find the exact area of R, giving your answer in the form ln(p/q), where p and q are positive integers.
The function f is defined for x > −1 and f′(x) = ln(x + 1) − 1. Figure 1 shows the graph of y = f′(x). It meets the y-axis at (0, −1) and the x-axis at the point P. The curve C has equation y = f(x) and passes through the point (0, 2).
(a)
Find the exact x-coordinate of P.
(2)
(b)
Explain what P tells you about the curve C, and use Figure 1 to state the nature of the corresponding point on C.
(2)
(c)
Use integration by parts to find ∫ln(x + 1) dx.
(3)
(d)
Hence find f(x) and the exact y-coordinate of the minimum point of C.
(3)
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Answer
Mark
Notes
Part (a)
ln(x + 1) = 1
M1
x = e − 1
A1
Part (b)
f′(x) = 0 at P, so C has a stationary point where x = e − 1
B1
f′(x) changes from negative to positive there, so it is a minimum
B1
Part (c)
writes as 1 × ln(x + 1): x ln(x + 1) − ∫x/(x + 1) dx
M1
x/(x + 1) = 1 − 1/(x + 1)
M1
(x + 1) ln(x + 1) − x + c (or x ln(x + 1) − x + ln(x + 1) + c)
A1
Part (d)
f(x) = (x + 1) ln(x + 1) − 2x + c
M1
f(0) = 2 gives c = 2
A1
y = e × 1 − 2(e − 1) + 2 = 4 − e
A1
Common error: Integrating f′(x) but forgetting the −1 term, which integrates to −x. f(x) needs both parts: the integral of ln(x + 1) and the integral of −1, plus the constant from (0, 2).
Figure 1 shows the curve with equation y = 8/(x2 + 4)3/2. The curve is symmetrical about the y-axis and has a maximum point at (0, 1). The shaded region R is bounded by the curve, the x-axis and the lines x = −2 and x = 2.
(a)
Use the substitution x = 2 tan θ to show that the area of R is given by ∫−π/4π/4 2 cos θ dθ.
(5)
(b)
Hence find the exact area of R.
(2)
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Answer
Mark
Notes
Part (a)
dx/dθ = 2 sec2 θ
B1
x2 + 4 = 4 tan2 θ + 4 = 4 sec2 θ
M1
(4 sec2 θ)3/2 = 8 sec3 θ
A1
8 × 2 sec2 θ/(8 sec3 θ) = 2/sec θ = 2 cos θ
M1
limits: x = ±2 gives tan θ = ±1, so θ = ±π/4
B1
Part (b)
[2 sin θ]−π/4π/4 = 2(√2/2) − 2(−√2/2)
M1
2√2
A1
Exam tip: Use the symmetry in Figure 1 as a check: the area from 0 to 2 is ∫0π/4 2 cos θ dθ = √2, exactly half of the total.
The curve C has equation y = (ln x)/x2, x > 0. The region R is bounded by C, the x-axis and the line x = e.
(a)
Write down the x-coordinate of the point where C meets the x-axis.
(1)
(b)
Use integration by parts to find ∫(ln x)/x2 dx.
(4)
(c)
Show that the area of R is 1 − 2/e.
(3)
(d)
Using integration by parts and your answer to part (c), find the exact value of ∫1e (ln x)2/x2 dx.
(4)
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Answer
Mark
Notes
Part (a)
ln x = 0, so x = 1
B1
Part (b)
u = ln x, dv/dx = x−2, so v = −x−1
B1
−(ln x)/x + ∫(1/x)(1/x) dx
M1
integrates x−2 correctly
M1
−(ln x)/x − 1/x + c
A1
Part (c)
uses limits 1 and e (from part (a)) in their answer to (b)
M1
(−1/e − 1/e) − (0 − 1)
M1
= 1 − 2/e (shown, with ln e = 1 and ln 1 = 0 seen)
A1
Part (d)
u = (ln x)2, du/dx = 2(ln x)/x, v = −1/x
B1
[−(ln x)2/x]1e + 2∫1e (ln x)/x2 dx
M1
−1/e + 2(1 − 2/e)
M1
2 − 5/e
A1
Model sketch
Figure: C meets the x-axis at (1, 0); R is shaded
Exam tip: In the last part the new integral after using parts is exactly the area from part (c). Spotting that a later integral reduces to an earlier one is what “using your answer to part (c)” is pointing you towards.
The region R is bounded by the curve y = x2/√(1 − x2), the x-axis and the line x = ½. The area of R is given by A = ∫0½x2/√(1 − x2) dx.
Figure: the region R; the curve has the asymptote x = 1
(a)
Use the substitution x = sin θ to show that A = ∫0π/6 sin2 θ dθ.
(5)
(b)
Hence find the exact value of A, giving your answer in the form π/a − √b/c, where a, b and c are integers.
(5)
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Answer
Mark
Notes
Part (a)
dx/dθ = cos θ, so dx = cos θ dθ
B1
√(1 − sin2 θ) = cos θ (valid since cos θ > 0 on the interval)
M1
sin2 θ × cos θ/cos θ = sin2 θ
A1
limits: x = 0 → θ = 0, x = ½ → θ = π/6
B1
fully correct integral with limits, A = ∫0π/6 sin2 θ dθ (shown)
A1
Part (b)
uses cos 2θ = 1 − 2 sin2 θ to write sin2 θ = ½ − ½ cos 2θ
M1
integrates to the form pθ + q sin 2θ
M1
½θ − ¼ sin 2θ
A1
(π/12 − ¼ sin(π/3)) − 0, with sin(π/3) = √3/2
M1
π/12 − √3/8
A1
Exam tip: sin² θ cannot be integrated directly at this level. Rewrite it with the double-angle identity first; this identity is not in the formulae booklet, so it must be learned.
Water leaks from a damaged storage tank. The rate of leaking, R litres per hour, t hours after the damage occurs is modelled by R = 20te−t/2, t ≥ 0. The tank holds far more water than will leak, so the model applies for all t ≥ 0.
(a)
Use integration by parts to find ∫te−t/2 dt.
(4)
(b)
Find the volume of water that leaks in the first 4 hours, giving your answer exactly and then to 3 significant figures.
(3)
(c)
Show that, according to the model, the total volume that leaks can never exceed 80 litres.
(2)
(d)
Find the percentage of this 80-litre limit that leaks in the first 4 hours.
(1)
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Answer
Mark
Notes
Part (a)
u = t, dv/dt = e−t/2, so v = −2e−t/2
B1
−2te−t/2 + ∫2e−t/2 dt
M1
∫2e−t/2 dt = −4e−t/2
M1
−2te−t/2 − 4e−t/2 + c
A1
Part (b)
20∫04te−t/2 dt = 20[−2te−t/2 − 4e−t/2]04
M1
20[(−8e−2 − 4e−2) − (0 − 4)] = 80 − 240e−2
A1
47.5 litres
A1
Part (c)
volume after T hours = 80 − (40T + 80)e−T/2
M1
(40T + 80)e−T/2 > 0 for all T ≥ 0, so the volume is always less than 80 litres (it tends to 80 as T → ∞)
A1
Part (d)
47.52… ÷ 80 × 100 = 59.4%
B1
Model sketch
Figure: the shaded area is the volume leaked in the first 4 hours
Common error: Integrating e−t/2 as −½e−t/2. Dividing by the coefficient −½ means multiplying by −2.
Figure 1 shows the curves C1 and C2 with equations y = ln x and y = (ln x)2 respectively, x > 0. The curves meet at the points A and B, where A lies on the x-axis. The shaded region R is bounded by C1 and C2 between A and B.
(a)
Find the exact coordinates of A and of B.
(3)
(b)
Use integration by parts to show that ∫(ln x)2 dx = x(ln x)2 − 2∫ln x dx.
(2)
(c)
Hence find ∫(ln x)2 dx.
(3)
(d)
Show that the area of R is 3 − e.
(4)
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Answer
Mark
Notes
Part (a)
ln x = (ln x)2, so ln x(ln x − 1) = 0
M1
A(1, 0)
A1
ln x = 1: B(e, 1)
A1
Part (b)
writes as 1 × (ln x)2 and differentiates (ln x)2 to 2 ln x/x
Exam tip: Read the order of the curves from Figure 1 before integrating. Between A and B, ln x lies between 0 and 1, so (ln x)2 is smaller and C1 is on top; the area is ∫(C1 − C2).
Figure 1 shows part of the curve C with equation y = e2x/(ex + 2)2, x ∈ ℝ. The line y = 1 is an asymptote to C. C meets the y-axis at A(0, 1/9) and passes through the point B(ln 4, 4/9). The horizontal line through B meets the y-axis at D. The shaded region R is bounded by C, the line segment AD and the line segment DB.
(a)
Use the substitution u = ex to show that ∫0ln 4e2x/(ex + 2)2 dx = ∫14u/(u + 2)2 du.
(4)
(b)
Express u/(u + 2)2 in the form A/(u + 2) + B/(u + 2)2, where A and B are constants.
(2)
(c)
Hence find the exact area of R, giving your answer in the form p + q ln 2, where p and q are rational numbers.
(5)
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Answer
Mark
Notes
Part (a)
du/dx = ex = u, so dx = du/u
B1
e2x = u2
B1
u2/(u + 2)2 × 1/u = u/(u + 2)2
M1
limits: x = 0 gives u = 1 and x = ln 4 gives u = 4
B1
Part (b)
u ≡ A(u + 2) + B, comparing coefficients or using u = −2
M1
1/(u + 2) − 2/(u + 2)2
A1
Part (c)
integrates 1/(u + 2) to a logarithm and 2/(u + 2)2 to a multiple of (u + 2)−1
M1
ln(u + 2) + 2/(u + 2) ft
A1
(ln 6 + 1/3) − (ln 3 + 2/3) = ln 2 − 1/3, the area under C from x = 0 to x = ln 4
A1
area under DB from x = 0 to x = ln 4: (4/9) × ln 4 = (8/9) ln 2
B1
R = (8/9) ln 2 − (ln 2 − 1/3) = 1/3 − (1/9) ln 2
A1
Common error: Integrating −2/(u + 2)2 as a logarithm. Only the power −1 gives a logarithm: (u + 2)−2 integrates to −(u + 2)−1, so −2/(u + 2)2 gives +2/(u + 2).
Figure 1 shows the curve C with equation y = (2 − x)ex, 0 ≤ x ≤ 2. C meets the y-axis at A(0, 2) and the x-axis at B(2, 0), and has a maximum point M. The shaded region R is bounded by C and the coordinate axes. Two trapezium strips of width 1 are also drawn.
(a)
Find the exact coordinates of M.
(3)
(b)
Use integration by parts to show that the area of R is e2 − 3.
(4)
(c)
The trapezium rule with the two strips shown is used to estimate the area of R. Find this estimate, giving your answer in terms of e.
(2)
(d)
Find the percentage error in this estimate, and explain, with reference to Figure 1, why the trapezium rule gives an underestimate here.
Figure 1 shows part of the curve C with equation y = 4(x − 3)/((x + 1)(x − 1)), x > 1. The line x = 1 is an asymptote to C. C crosses the x-axis at Q. The shaded region R1 is bounded by C, the x-axis and the line x = 2. The shaded region R2 is bounded by C, the x-axis and the line x = 7.
Find the exact area of R1 and the exact area of R2, each in the form k ln(p/q).
(5)
(d)
Hence find ∫27y dx and the total area of the shaded regions, each in the form k ln(p/q).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
4(x − 3) ≡ A(x − 1) + B(x + 1)
M1
one constant correct
A1
8/(x + 1) − 4/(x − 1)
A1
Part (b)
(3, 0)
B1
Part (c)
integrates to logarithms
M1
8 ln(x + 1) − 4 ln(x − 1)
A1
∫23y dx = (8 ln 4 − 4 ln 2) − 8 ln 3 = 4 ln(8/9), negative as R1 is below the axis, so R1 = 4 ln(9/8)
A1
∫37y dx = (8 ln 8 − 4 ln 6) − (8 ln 4 − 4 ln 2)
M1
R2 = 4 ln(4/3)
A1
Part (d)
∫27y dx = 4 ln(4/3) − 4 ln(9/8) = 4 ln(32/27) ft
B1
total area = 4 ln(4/3) + 4 ln(9/8) = 4 ln(3/2) ft
B1
Exam tip: Use Figure 1 to split the integral at Q. A single integral from 2 to 7 subtracts the area of R1 from the area of R2; it does not give the total shaded area.
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