Integration: exam review Exam review: second paper Worked solutions

The graph shows the speed of a cyclist on a track during a journey of 22 s. It speeds up steadily from rest for the first 6 s, then travels at 9 m/s until 15 s, then slows steadily to rest. [5]

1. (a) Find the distance travelled in the first 6 s. [2]

   • M1 for the area of the triangle: \(\frac{1}{2}\times 6\times 9\)
   • A1 27 m

2. (b) Find the total distance travelled. [3]

   • M1 for identifying the shapes: triangle, rectangle and triangle, or a trapezium with parallel sides 22 and 9
   • M1 for correct areas or substitution: \(\frac{1}{2}(22+9)\times 9\) or \(27+81+31.5\)
   • A1 139.5 m

Find each indefinite integral. [7]

1. (a) \(\displaystyle\int\left(8x^{3} - \frac{15}{x^{4}} + 6\right)dx\) [3]

   • M1 for adding 1 to a power and dividing by the new power
   • A1 for at least two terms correct
   • A1 \(2x^{4} + 6x + \frac{5}{x^{3}} + c\), including \(+c\)

2. (b) \(\displaystyle\int (x+2)(3x-1)\,dx\) [2]

   • M1 for expanding the brackets: \(3x^{2} + 5x - 2\)
   • A1 \(x^{3} + \frac{5}{2}x^{2} - 2x + c\)

3. (c) \(\displaystyle\int \frac{12x^{4}-5x^{2}+9}{x^{2}}\,dx\) [2]

   • M1 for splitting the fraction: \(12x^{2} - 5 + \frac{9}{x^{2}}\)
   • A1 \(4x^{3} - 5x - \frac{9}{x} + c\)

Show each integration step. [6]

1. (a) Find \(\displaystyle\int\left(9\sqrt{x} + \frac{10}{\sqrt{x}}\right)dx\). [3]

   • M1 for adding 1 to a power and dividing by the new power
   • A1 for at least one term correct
   • A1 \(6x^{\frac{3}{2}} + 20\sqrt{x} + c\), including \(+c\)

2. (b) A curve has gradient \(\dfrac{dy}{dx}=6\sqrt{x} - 7\) and passes through the point \((9, 20)\). Find the equation of the curve. [3]

   • M1 for integrating: \(y=4x^{\frac{3}{2}} - 7x + c\)
   • M1 for substituting \(x=9\), \(y=20\): \(20=45+c\)
   • A1 \(y=4x^{\frac{3}{2}} - 7x - 25\)

Show your working. [7]

1. (a) Find \(\displaystyle\int \frac{7}{x^{4}}\,dx\). [2]

   • M1 for writing \(7x^{-4}\) and raising the power by 1, then dividing by the new power
   • A1 \(-\dfrac{7}{3x^{3}}+c\), including \(+c\)

2. (b) Expand \(\left(x^{3}-\dfrac{2}{x}\right)^{2}\) and hence find \(\displaystyle\int \left(x^{3}-\frac{2}{x}\right)^{2}dx\). [3]

   • M1 for expanding: \(x^{6}-4x^{2}+4x^{-2}\) (A1 for all three terms correct)
   • A1 \(x^{6}-4x^{2}+\dfrac{4}{x^{2}}\)
   • A1 \(\dfrac{x^{7}}{7}-\dfrac{4x^{3}}{3}-\dfrac{4}{x}+c\)

3. (c) Given that \(\displaystyle\int g(x)\,dx=3x^{4}-\frac{8}{x^{3}}+c\), find \(g(x)\). [2]

   • M1 for differentiating the given expression term by term: \(3x^{4}-8x^{-3}\)
   • A1 \(12x^{3}+24x^{-4}\)

The profile of a ski jump is modelled by a curve, with \(x\) and \(y\) in metres. The gradient of the curve is \(\dfrac{dy}{dx}=12x^{2} + 3 - \frac{9}{x^{4}}\). The curve passes through the point \((1, 14)\). [5]

1. (a) Find the equation of the curve, in the form \(y=f(x)\). [4]

   • M1 for integrating the gradient function
   • A1 \(y=4x^{3} + 3x + \frac{3}{x^{3}} + c\)
   • M1 for substituting \(x=1\), \(y=14\) to find \(c\): \(c=4\)
   • A1 \(y=4x^{3} + 3x + \frac{3}{x^{3}} + 4\)

2. (b) Find the value of \(y\) on the curve when \(x=2\). [1]

   • A1 \(y=42.4\) (ft from part (a))

A function \(f\) has gradient function \(f'(x)=4x^{3}-6x\), and \(f(0)=4\). [5]

1. (a) Find \(f(x)\). [3]

   • M1 for integrating \(f'(x)\) (A1 for the two terms correct)
   • A1 \(f(x)=x^{4}-3x^{2}+4\), using \(f(0)=4\) to find \(c\)

2. (b) Hence find \(\displaystyle\int_0^{2} f(x)\,dx\). [2]

   • M1 for writing \(\int_0^{2}\left(x^{4}-3x^{2}+4\right)dx\)
   • A1 6.4, from the calculator

The curve \(y=x^{2}-8x+18\) is shown. The region under the curve between \(x=2\) and \(x=6\) is shaded. [6]

1. (a) Write down a definite integral for the area of the shaded region. [2]

   • M1 for an integral of the curve with the correct limits
   • A1 \(\displaystyle\int_{2}^{6}\left(x^{2} - 8x + 18\right)dx\)

2. (b) Use your calculator to evaluate the integral. Give your answer to 3 significant figures. [2]

   • M1 for using the calculator on the integral
   • A1 13.3

3. (c) Use your calculator to evaluate \(\displaystyle\int_{1}^{3}\left(3x + \frac{10}{x^{2}}\right)dx\). Give your answer to 3 significant figures. [2]

   • M1 for using the calculator with the integrand rewritten with powers, or entered as given
   • A1 18.7

For the first 30 s after a gate is lifted, a sluice releases water at a rate directly proportional to the time since it was lifted. After 4 s the rate is 36 litres per second. [4]

1. (a) Find an expression for the rate \(r\) litres per second, at time \(t\) seconds. [2]

   • M1 for \(r=ct\) and substituting: \(36=c\times4\)
   • A1 \(r=9t\)

2. (b) Find the amount of water released in the first 30 seconds. [2]

   • M1 for \(\displaystyle\int_0^{30}9t\,dt\)
   • A1 4050 litres, from the calculator

The rate at which the value of a rare coin changes is modelled by \(y=2e^{x}\) euros per year, where \(x\) is the time in years. The total change from \(x=1\) to \(x=3\) is \(\displaystyle\int_{1}^{3} 2e^{x}\,dx\). Use the trapezium rule with 4 strips to estimate this total. [7]

1. (a) Write down the values of \(y=2e^{x}\) at \(x=1.5\), \(x=2\) and \(x=2.5\), correct to 4 significant figures. [2]

   • A1 for two values correct
   • A1 8.963, 14.78, 24.36

2. (b) Use the trapezium rule to find the estimate. Give your answer to 3 significant figures. [3]

   • M1 for \(h=0.5\)
   • M1 for the rule: \(\frac{0.5}{2}\left[5.437+40.17+2(8.963+14.78+24.36)\right]\)
   • A1 35.5

3. (c) State whether the estimate is greater or less than the exact value of the integral. Give a reason. [2]

   • A1 greater than (an overestimate)
   • R1 for a reason: the curve is concave up (bends upwards), so the straight tops of the trapeziums are above it

The map shows a reservoir. The top and bottom edges of the reservoir are marked at equal intervals along the \(x\)-axis. Points on the top edge are (0, 70), (12, 92), (24, 85), (36, 96), (48, 64). Points on the bottom edge are (0, 20), (12, 15), (24, 22), (36, 11), (48, 18). Distances are in metres. [5]

1. (a) Use the trapezium rule to estimate the area between the top edge and the x-axis, from \(x=0\) to \(x=48\). [2]

   • M1 for the rule with \(h=12\): \(\frac{12}{2}\left[70+64+2(92+85+96)\right]\)
   • A1 4080

2. (b) Use the trapezium rule to estimate the area between the bottom edge and the x-axis, over the same interval. [2]

   • M1 for the rule with \(h=12\): \(\frac{12}{2}\left[20+18+2(15+22+11)\right]\)
   • A1 804

3. (c) Hence estimate the area of the reservoir. Give your answer to 3 significant figures. [1]

   • A1 \(4080-804=3276\), so 3280 m² to 3 significant figures (ft)

The function \(f(x)=x^{3}+2x\), for \(x\ge0\), is one-to-one, with inverse function \(f^{-1}\). [4]

1. (a) Use your calculator to find \(\displaystyle\int_0^{2} f(x)\,dx\). [1]

   • A1 8

2. (b) By considering a sketch of \(y=f(x)\) and a rectangle, find \(\displaystyle\int_0^{12} f^{-1}(x)\,dx\). [3]

   • A1 \(f(2)=12\)
   • M1 for the rectangle with corners \((0,0)\) and \((2,12)\), area \(2\times12\), less the area from (a)
   • A1 16

Fuel is pumped into a storage drum at a rate of \(r\) litres per minute, where \(r=kt+1.5\) and \(t\) is the time in minutes after the pump is switched on, for \(0\le t\le 10\). The graph of \(r\) against \(t\) is shown. When \(t=6\), \(r=6.3\). At \(t=0\) the drum already holds 30 litres. [12]

1. (a) Find the value of \(k\). [2]

   • M1 for substituting \(t=6\), \(r=6.3\): \(6k+1.5=6.3\)
   • A1 \(k=0.8\)

2. (b) Find the amount in the drum when \(t=10\). [3]

   • M1 for the integral of the rate: \(\displaystyle\int_0^{10}\left(0.8t+1.5\right)dt\)
   • M1 for adding the starting amount: \(30+\) integral
   • A1 85 litres

3. (c) Find the value of \(t\) when the drum holds 70 litres. [3]

   • M1 for an equation: \(30+\displaystyle\int_0^{T}\left(0.8t+1.5\right)dt=70\)
   • M1 for solving on the calculator (Solve for T)
   • A1 \(T=8.30\) minutes

4. (d) The shaded region on the graph of \(r\) is the area between \(t=3\) and \(t=7\). Write down what this area represents, and find its value. [3]

   • A1 the amount that flows in between \(t=3\) and \(t=7\) (litres)
   • M1 for \(\displaystyle\int_{3}^{7}\left(0.8t+1.5\right)dt\)
   • A1 22 litres

5. (e) Give one criticism of this model. [1]

   • A1 for a sensible criticism, for example the rate increases without limit, or the model only applies for \(0\le t\le 10\)

A train pulls out of a depot. The speed, \(v\) m/s, is modelled by \(v=4\sqrt{t}\) for \(0\le t\le 36\), where \(t\) is the time in seconds. Values of \(v\) are given to 4 significant figures. Time \(t\) (s): 0, 9, 18, 27, 36. Speed \(v\) (m/s): 0, 12, 16.97, 20.78, 24. [12]

1. (a) State what the area under the graph of \(v\) against \(t\) represents. [1]

   • A1 the distance travelled (accept: the distance from the start to \(t=36\))

2. (b) Use the trapezium rule with 4 strips and the values in the table to estimate the distance travelled in the 36 seconds. Give your answer to 3 significant figures. [3]

   • M1 for \(h=9\)
   • M1 for the rule: \(\frac{9}{2}\left[0+24+2(12+16.97+20.78)\right]\)
   • A1 556 m

3. (c) Is your estimate greater or less than the true distance? Give a reason from the shape of the graph. [2]

   • A1 less than (an underestimate)
   • R1 for a reason: the graph is concave down (bends downwards), so the straight tops of the trapeziums are below it

4. (d) Use your calculator to find \(\displaystyle\int_0^{36} 4\sqrt{t}\,dt\). Give your answer to 3 significant figures. [2]

   • M1 for using the calculator with \(\displaystyle\int_0^{36}4\sqrt{t}\,dt\)
   • A1 576 m

5. (e) Find the percentage error of the trapezium estimate. Give your answer to 3 significant figures. [2]

   • M1 for \(\frac{|\text{estimate}-\text{exact}|}{\text{exact}}\times100\): \(\frac{576-555.75}{576}\times100\)
   • A1 3.52%

6. (f) Find the time at which the train has travelled 100 m from the start. Give your answer to 3 significant figures. [2]

   • M1 for the equation \(\displaystyle\int_0^{T}4\sqrt{t}\,dt=100\) solved on the calculator
   • A1 \(T=11.2\) s

A reservoir is dug in a park. Its boundary is part of the curve \(y=12x-x^{2}\) and part of the line \(y=4x\), with \(x\) and \(y\) in metres. The reservoir is the region shaded in the diagram. [12]

1. (a) Find the \(x\)-coordinate of the point where the curve meets the \(x\)-axis, other than the origin. [1]

   • A1 \(x=12\) from \(12x-x^2=0\)

2. (b) Find the \(x\)-coordinate of the point P, where the curve meets the line, other than the origin. [2]

   • M1 for equating the curve and the line: \(12x-x^{2}=4x\)
   • A1 \(x=8\)

3. (c) Write down an integral for the area of the shaded region. [2]

   • M1 for curve minus line as the integrand
   • A1 \(\displaystyle\int_0^{8}\left(8x-x^{2}\right)dx\) with correct limits

4. (d) Use your calculator to find the area of the shaded region. [2]

   • M1 for using the calculator on the integral
   • A1 85.3 m² (exactly \(\frac{256}{3}\))

5. (e) Find the area of the region between the curve and the \(x\)-axis. [2]

   • M1 for \(\displaystyle\int_0^{12}\left(12x-x^{2}\right)dx\)
   • A1 288 m²

6. (f) Find the percentage of the area in part (e) that the shaded region makes up. Give your answer to 3 significant figures. [2]

   • M1 for \(\frac{85.33}{288}\times100\)
   • A1 29.6%

7. (g) The shaded region has a constant depth of 0.8 m. Find its volume. [1]

   • A1 85.33\(\times 0.8\) = 68.3 m³ (ft)

A vegetable plot is the region between the curve \(y=\dfrac{24}{x}\) and the line \(y=13-x\), with \(x\) and \(y\) in metres. The curve and the line meet at the points P and Q. [12]

1. (a) Use your calculator to find the \(x\)-coordinates of P and Q. [2]

   • M1 for equating: \(\dfrac{24}{x}=13-x\)
   • A1 \(x=2.23\) and \(x=10.8\)

2. (b) Write down an integral for the area of the vegetable plot, and use your calculator to find the area. [3]

   • M1 for the line minus the curve as the integrand (the line is above the curve between P and Q)
   • A1 \(\displaystyle\int_{2.228}^{10.772}\left((13-x)-\frac{24}{x}\right)dx\) with correct limits
   • A1 17.7 m²

3. (c) Gravel is laid on the ground under the curve, between \(x=3\) and \(x=11\). Use the trapezium rule with 4 strips to estimate the area of this ground. [3]

   • M1 for \(h=2\) and the five y-values (x: 3, 5, 7, 9, 11; y: 8, 4.8, 3.429, 2.667, 2.182)
   • M1 for the trapezium rule: \(\tfrac12(2)[(8+2.182)+2(4.8+3.429+2.667)]=31.97\)
   • A1 32.0 m²

4. (d) Use your calculator to find the exact value of \(\displaystyle\int_{3}^{11}\frac{24}{x}\,dx\). [1]

   • A1 31.2, from the calculator

5. (e) Find the percentage error in your estimate in part (c). [2]

   • M1 for \(\dfrac{31.97-31.18}{31.18}\times100\)
   • A1 2.53%

6. (f) The vegetable plot has a constant depth of 0.5 m. Find its volume. [1]

   • A1 8.86 m³ (ft)