Integration Applications Paper 1 Review [114 marks]
1.
Let f ′(x) = sin 3(2x) cos(2x). Find f(x), given that f ( π ) = 1.
[7 marks]
4
Markscheme
evidence of integration
eg ∫
(M1)
f ′(x)dx
correct integration (accept missing C)
eg
1
2
×
sin4(2x)
(A2)
, 18 sin4(2x) + C
4
substituting initial condition into their integrated expression (must have +C)
eg 1 =
1
sin4 ( π2 )
8
M1
+C
Note: Award M0 if they substitute into the original or differentiated function.
recognizing sin( π ) = 1
(A1)
2
eg 1 = 18 (1)4 + C
C=
7
8
(A1)
f(x) = 18 sin4(2x) + 78
A1
N5
[7 marks]
Let f ′(x) =
6− 2x
,
6x−x2
for 0 < x < 6.
The graph of
f has a maximum point at P.
2a. Find the x-coordinate of P.
[3 marks]
Markscheme
recognizing f ′(x) = 0
correct working
(M1)
(A1)
eg 6 − 2x = 0
x=3
A1
N2
[3 marks]
The
y-coordinate of P is ln 27.
2b. Find f(x), expressing your answer as a single logarithm.
[8 marks]
Markscheme
evidence of integration
eg ∫
f ′,
∫
using substitution
eg ∫
1
u du
(M1)
6− 2x
dx
6x−x2
(A1)
where u = 6x − x2
correct integral
A1
eg ln(u) + c, ln(6x − x2)
substituting (3, ln 27) into their integrated expression (must have c)
eg
ln(6 × 3 − 32) + c
correct working
(M1)
= ln 27, ln(18 − 9) + ln k = ln 27
(A1)
eg c = ln 27 − ln 9
EITHER
c = ln 3
(A1)
attempt to substitute their value of c into f(x)
eg f(x) =
ln(6x − x2) + ln 3
A1
(M1)
N4
OR
attempt to substitute their value of c into f(x)
eg f(x) =
(M1)
ln(6x − x2) + ln 27 − ln 9
correct use of a log law
(A1)
eg f(x) = ln(6x − x2) + ln( 27 ), f(x) = ln(27(6x − x2)) − ln 9
9
f(x) =
ln(3(6x − x2))
A1
N4
[8 marks]
2c.
The graph of
f is transformed by a vertical stretch with scale factor
Find the value of a and of b, where a, b ∈ N .
Markscheme
a=3
A1
N1
correct working
eg
A1
ln 27
ln 3
correct use of log law
eg
3 ln 3
,
ln 3
b=3
log327
A1
[4 marks]
N2
(A1)
1
.
ln 3
The image of P under this transformation has coordinates (a, b).
The following diagram shows the graph of f(x) = 2x√a2 − x2, for
−1 ⩽ x ⩽ a, where a > 1.
The line L is the tangent to the graph of f at the origin, O. The point
P(a, b) lies on L.
Given that f ′(x) =
3a. (i)
2a2− 4x2
,
√a2−x2
for −1 ⩽ x < a, find the equation of
[6 marks]
L.
(ii)
Hence or otherwise, find an expression for b in terms of a.
Markscheme
(i)
eg
recognizing the need to find the gradient when x = 0 (seen anywhere)
f ′(0)
correct substitution
f ′(0) =
f ′(0)
(A1)
2a2− 4(0)
√a2−0
= 2a (A1)
correct equation with gradient 2
a (do not accept equations of the form L = 2ax)
A1
N3
eg y = 2ax, y − b = 2a(x − a), y = 2ax − 2a2 + b
METHOD 1
(ii)
attempt to substitute x = a into their equation of L
eg y = 2a × a
b = 2a2
A1
N2
METHOD 2
equating gradients
eg
b
a
= 2a
b = 2a2 A1
[6 marks]
N2
(M1)
(M1)
R1
The point
Q(a, 0) lies on the graph of f. Let R be the region enclosed by the graph of f and the x-axis. This information is shown in the following
diagram.
Let AR be the area of the region R.
2 3
3b. Show that AR = 3 a .
[6 marks]
Markscheme
METHOD 1
a
recognizing that area = ∫ 0 f(x)dx (seen anywhere)
valid approach using substitution or inspection
eg ∫ 2x√udx, u = a2 − x2, du = −2xdx,
correct working
R1
(M1)
3
2 2
(a − x2) 2
3
(A1)
eg ∫ 2x√a2 − x2dx = ∫ −√udu
3
∫ −√udu = − u32
(A1)
2
3
∫ f(x)dx = − 23 (a2 − x2) 2 + c
(A1)
substituting limits and subtracting
eg AR =
3
− 23 (a2 − a2) 2
AR = 23 a3 AG
+
A1
3
3
2 2
(a − 0) 2 , 23 (a2) 2
3
N0
METHOD 2
a
recognizing that area = ∫ 0 f(x)dx (seen anywhere)
valid approach using substitution or inspection
eg ∫ 2x√udx, u = a2 − x2, du = −2xdx,
correct working
(M1)
3
2 2
(a − x2) 2
3
(A1)
eg ∫ 2x√a2 − x2dx = ∫ −√udu
3
∫ −√udu = − u32
(A1)
2
new limits for u (even if integration is incorrect)
eg u = 0 and u = a2, ∫ 0 u 2 du, [− 23 u 2 ]
a2
1
3
substituting limits and subtracting
eg AR = − (0 − 2 a3) ,
3
AR = 23 a3 AG
[6 marks]
N0
2 2
(a )
3
3
2
A1
0
a2
R1
(A1)
3c. Let AT be the area of the triangle OPQ. Given that AT = kAR, find the value of k.
Markscheme
METHOD 1
valid approach to find area of triangle
eg
(M1)
1
(OQ)(PQ), 1 ab
2
2
correct substitution into formula for AT (seen anywhere)
eg AT =
1
2
× a × 2a2,
a3
valid attempt to find k (must be in terms of a)
eg
k=
a3
=
3
2
k 23 a3,
k=
A1
N2
(M1)
a3
2 3
a
3
METHOD 2
valid approach to find area of triangle
eg
correct working
eg
(M1)
a
∫ 0 (2ax)dx
[ax2]a0 ,
(A1)
a3
valid attempt to find k (must be in terms of a)
eg
k=
a3
3
2
=
k 23 a3,
k=
A1
N2
[4 marks]
a3
2 3
a
3
(M1)
(A1)
[4 marks]
Let y = f(x), for
−0.5 ≤ x
≤
6.5. The following diagram shows the graph of f ′, the derivative of f.
The graph of f ′ has a local maximum when x = 2, a local minimum when x = 4, and it crosses the
x-axis at the point
(5, 0).
4a. Explain why the graph of f has a local minimum when x = 5.
[2 marks]
Markscheme
METHOD 1
f ′(5) = 0
(A1)
valid reasoning including reference to the graph of f ′
R1
eg f ′ changes sign from negative to positive at x = 5, labelled sign chart for f ′
so f has a local minimum at x = 5
AG
N0
Note: It must be clear that any description is referring to the graph of f ′, simply giving the conditions for a minimum without relating
them to f ′ does not gain the R1.
METHOD 2
f ′(5) = 0
A1
valid reasoning referring to second derivative
eg
f ′′(5)
R1
>0
so f has a local minimum at x = 5
AG
N0
[2 marks]
4b. Find the set of values of x for which the graph of f is concave down.
[2 marks]
Markscheme
attempt to find relevant interval
eg
f′
(M1)
is decreasing, gradient of f ′ is negative, f ′′ < 0
2 < x < 4 (accept “between 2 and 4”)
Notes:
A1
N2
If no other working shown, award M1A0 for incorrect inequalities such as 2 ≤ x ≤ 4, or “from 2 to 4”
[2 marks]
4c. The following diagram shows the shaded regions A, B and C.
The regions are enclosed by the graph of f ′, the x-axis, the y-axis, and the line x = 6.
The area of region A is 12, the area of region B is 6.75 and the area of region C is 6.75.
Given that f(0) = 14, find f(6).
[5 marks]
Markscheme
METHOD 1 (one integral)
correct application of Fundamental Theorem of Calculus
eg
6
∫ 0 f ′(x)dx
=f(6) − f(0), f(6) =
attempt to link definite integral with areas
eg
6
∫ 0 f ′(x)dx
correct value for
= −12 − 6.75 + 6.75,
6
∫ 0 f ′(x)dx
(A1)
6
14 + ∫ 0 f ′(x)dx
(M1)
6
∫ 0 f ′(x)dx
= Area A + Area B + Area C
(A1)
6
eg ∫ 0 f ′(x)dx = −12
correct working
A1
eg f(6) − 14 = −12, f(6) = −12 + f(0)
f(6) = 2
A1
N3
METHOD 2 (more than one integral)
correct application of Fundamental Theorem of Calculus
eg
∫ 02 f ′(x)dx
= f(2) − f(0), f(2) =
attempt to link definite integrals with areas
eg
∫ 02 f ′(x)dx
= 12,
∫ 25 f ′(x)dx
correct values for integrals
eg
∫ 02 f ′(x)dx
= −12,
[5 marks]
A1
N3
6
(A1)
one correct intermediate value
f(6) = 2
(M1)
= −6.75, ∫ 0 f ′(x) = 0
∫ 52 f ′(x)dx
eg f(2) = 2, f(5) = −4.75
(A1)
14 + ∫ 02 f ′(x)
A1
= 6.75, f(6) − f(2) = 0
4d.
The following diagram shows the shaded regions A, B and C.
[6 marks]
The regions are enclosed by the graph of f ′, the x-axis, the y-axis, and the line x = 6.
The area of region A is 12, the area of region B is 6.75 and the area of region C is 6.75.
Let g(x) = (f(x))2. Given that f ′(6) = 16, find the equation of the tangent to the graph of g at the point where x = 6.
Markscheme
correct calculation of g(6) (seen anywhere)
A1
eg 22, g(6) = 4
choosing chain rule or product rule
eg g ′ (f(x)) f ′(x),
dy
dx
correct derivative
(A1)
=
dy
du
×
du
,
dx
(M1)
f(x)f ′(x) + f ′(x)f(x)
eg g ′(x) = 2f(x)f ′(x), f(x)f ′(x) + f ′(x)f(x)
correct calculation of g ′(6) (seen anywhere)
A1
eg 2(2)(16), g ′(6) = 64
attempt to substitute their values of g ′(6) and g(6) (in any order) into equation of a line
eg
22
(M1)
= (2 × 2 × 16)6 + b, y − 6 = 64(x − 4)
correct equation in any form
A1
N2
eg y − 4 = 64(x − 6), y = 64x − 380
[6 marks]
[Total 15 marks]
5.
Let f ′(x) = 6x2 − 5. Given that f(2) = −3, find f(x).
[6 marks]
Markscheme
evidence of antidifferentiation
eg f = ∫
(M1)
f′
correct integration (accept absence of C)
f(x) =
6x3
3
− 5x + C,
(A1)(A1)
2x3 − 5x
attempt to substitute (2, − 3) into their integrated expression (must have C)
eg
2(2)3 − 5(2) + C
Note:
M1
= −3, 16 − 10 + C = −3
Award M0 if substituted into original or differentiated function.
correct working to find C
(A1)
eg 16 − 10 + C = −3, 6 + C = −3, C = −9
f(x) = 2x3 − 5x − 9
A1
N4
[6 marks]
6.
Let f(x) = cos x, for 0 ≤ x ≤ 2π. The following diagram shows the graph of f.
There are
x-intercepts at x = π ,
2
3π
.
2
The shaded region R is enclosed by the graph of f, the line x = b, where b >
(1 −
√3
2
[8 marks]
) . Find the value of b.
3π
,
2
and the x-axis. The area of R is
Markscheme
attempt to set up integral (accept missing or incorrect limits and missing dx)
b
eg ∫ 3π cos xdx,
2
b
∫ a cos xdx,
M1
b
∫ 3π fdx, ∫ cos x
2
correct integration (accept missing or incorrect limits)
(A1)
eg [sin x]b3π , sin x
2
substituting correct limits into their integrated function and subtracting (in any order)
(M1)
eg sin b − sin( 3π ), sin( 3π ) − sin b
2
sin(
3π
)
2
2
= −1 (seen anywhere)
(A1)
setting their result from an integrated function equal to (1 −
eg sin b = −
√3
2
)
M1
√3
2
evaluating sin − 1 (
√3
2
)=
π
3
or sin − 1 (−
√3
2
) = − π3
(A1)
eg b = 3 , − 60∘
π
identifying correct value
eg 2π −
b=
5π
3
π
,
3
360 − 60
A1
N3
(A1)
[8 marks]
7.
The following diagram shows the graph of f(x) =
x
,
x2+ 1
for 0 ≤ x ≤ 4, and the line x = 4.
Let R be the region enclosed by the graph of f , the x-axis and the line x = 4.
Find the area of R.
Markscheme
substitution of limits or function
4
eg A = ∫ 0 f(x), ∫
(A1)
x
dx
x2+ 1
correct integration by substitution/inspection
A2
1
ln(x2 + 1)
2
substituting limits into their integrated function and subtracting (in any order)
eg
correct working
eg
(M1)
1
(ln(42 + 1) − ln(02 + 1))
2
A1
1
(ln(42 + 1) − ln(02 + 1)) , 1 (ln(17) − ln(1)) , 1 ln 17 − 0
2
2
2
A = 1 ln(17)
2
Note:
A1
N3
Exception to FT rule. Allow full FT on incorrect integration involving a ln function.
[6 marks]
[6 marks]
The following diagram shows the graph of a function f. There is a local minimum point at A, where x > 0.
The derivative of f is given by f ′(x) = 3x2 − 8x − 3.
8a.
Find the x-coordinate of A.
[5 marks]
Markscheme
recognizing that the local minimum occurs when f ′(x) = 0
valid attempt to solve 3x2 − 8x − 3 = 0
(M1)
(M1)
eg factorization, formula
correct working
A1
(3x + 1)(x − 3), x =
8± √64+36
6
x=3
A2
Note:
Award A1 if both values x =
N3
−1
,
3
x = 3 are given.
[5 marks]
8b. The y-intercept of the graph is at ( 0,6). Find an expression for f(x).
[6 marks]
m
The graph of a function g is obtained by reflecting the graph of f in the y-axis, followed by a translation of ( ).
n
Markscheme
valid approach
(M1)
f(x) = ∫ f ′(x)dx
f(x) = x3 − 4x2 − 3x + c (do not penalize for missing “+c”)
c=6
A1A1A1
(A1)
f(x) = x3 − 4x2 − 3x + 6
A1
N6
[6 marks]
Let
f(x) = x2 .
9a.
Find
∫ 12 (f(x))2dx.
[4 marks]
Markscheme
substituting for
(f(x))2 (may be seen in integral)
A1
eg
(x2)2, x4
correct integration,
∫ x4dx = 1 x5 (A1)
5
substituting limits into their integrated function and subtracting (in any order)
(M1)
eg
25
5
− 15 , 15 (1 − 4)
∫ 12 (f(x))2dx =
31
(=
5
6.2)
A1
N2
[4 marks]
9b. The following diagram shows part of the graph of
f.
The shaded region
R is enclosed by the graph of
f, the
x-axis and the lines
x = 1 and
x = 2.
Find the volume of the solid formed when
R is revolved
360∘ about the
x-axis.
Markscheme
attempt to substitute limits or function into formula involving
f 2 (M1)
eg
∫ 12 (f(x))2dx, π ∫ x4dx
31
π
5
(= 6.2π)
[2 marks]
A1
N2
[2 marks]
10.
Let
[7 marks]
a
∫ π cos 2xdx = 12 , where π < a < 2π. Find the value of
a.
Markscheme
correct integration (ignore absence of limits and “
+C”) (A1)
eg
sin(2x)
2
, ∫ π cos 2x = [ 12 sin(2x)]
a
a
π
substituting limits into their integrated function and subtracting (in any order)
(M1)
eg
1
sin(2a) − 12 sin(2π),
2
sin(2π) = 0
sin(2π) − sin(2a)
(A1)
setting their result from an integrated function equal to
1
M1
2
eg
1
sin 2a
2
= 12 , sin(2a) = 1
recognizing
sin− 11 =
π
2
(A1)
eg
2a = π2 , a =
π
4
correct value
(A1)
eg
π
2
+ 2π, 2a =
a=
5π
4
5π
,
2
A1
a=
π
4
+π
N3
[7 marks]
Let
f(x) =
11a.
2x
.
x2 +5
Use the quotient rule to show that
f ′(x) =
10− 2x2
(x2+5)2
[4 marks]
.
Markscheme
derivative of
2x is
2 (must be seen in quotient rule)
(A1)
derivative of
x2 + 5 is
2x (must be seen in quotient rule)
correct substitution into quotient rule
(A1)
A1
eg
(x2+5)(2)−(2x)(2x)
(x2+5)2
,
2(x2+5)− 4x2
(x2+5)2
correct working which clearly leads to given answer A1
eg
2x2+10− 4x2
(x2+5)2
f ′(x) =
,
2x2+10− 4x2
x4+10x2+25
10− 2x2
(x2+5)2
[4 marks]
AG
N0
11b.
Find
∫
[4 marks]
2x
dx.
x2+5
Markscheme
valid approach using substitution or inspection
(M1)
eg
u = x2 + 5, du = 2xdx, 1 ln(x2 + 5)
2
∫
2x
dx
x2+5
= ∫ u1 du
∫ u1 du = ln u + c
ln(x2 + 5) + c
(A1)
(A1)
A1
N4
[4 marks]
11c.
The following diagram shows part of the graph of
f.
The shaded region is enclosed by the graph of
f, the
x-axis, and the lines
x = √5 and
x = q. This region has an area of
ln 7. Find the value of
q.
[7 marks]
Markscheme
correct expression for area
(A1)
eg
q
∫
x
[ln(x2 + 5)]q√5, √5 x22+5
dx
substituting limits into their integrated function and subtracting (in either order)
(M1)
eg
2
ln(q 2 + 5) − ln(√5 + 5)
correct working
(A1)
eg
ln(q 2 + 5) − ln 10, ln
q2+5
10
equating their expression to
ln 7 (seen anywhere) (M1)
eg
ln(q 2 + 5) − ln 10 = ln 7, ln
correct equation without logs
q2+5
10
= ln 7, ln(q 2 + 5) = ln 7 + ln 10
(A1)
eg
q2+5
10
= 7, q 2 + 5 = 70
q 2 = 65
(A1)
q = √65
A1
N3
Note: Award A0 for
q = ±√65.
[7 marks]
12.
The graph of a function h passes through the point
( π ,5) .
12
Given that
h′(x) = 4 cos 2x, find
h(x).
[6 marks]
Markscheme
evidence of anti-differentiation
(M1)
eg
∫ h′(x),∫ 4 cos 2xdx
correct integration
(A2)
eg
h(x) = 2 sin 2x + c,
4 sin 2x
2
attempt to substitute
( π ,5) into their equation
12
(M1)
eg
2 sin(2 × 12π ) + c = 5, 2 sin( π6 ) = 5
correct working
(A1)
eg
2 ( 12 ) + c = 5, c = 4
h(x) = 2 sin 2x + 4
A1
N5
[6 marks]
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