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IB Mathematics: Applications and Interpretation HL · AHL 3: Further Integration · AHL 3.14

Further Integration: exam review Exam review One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Section A: short questions

  1. Find the following. [6]

    1. (a) (a) \(\int\left(6\sqrt{x}-\frac{10}{\sqrt{x}}\right)dx\) [3]

      • M1 for rewriting as \(6x^{\frac{1}{2}}-10x^{-\frac{1}{2}}\)
      • A1 for one correct term
      • A1 for \(4x^{\frac{3}{2}}-20x^{\frac{1}{2}}+c\) (including \(+c\))

    2. (b) (b) Given that \(\frac{dy}{dx}=6e^{x}-3\cos x\) and that \(y=4\) when \(x=0\), find \(y\) in terms of \(x\). [3]

      • M1 for integrating at least one term correctly
      • A1 for \(6e^{x}-3\sin x+c\)
      • A1 for \(c=-2\) and \(y=6e^{x}-3\sin x-2\)

  2. Find each integral. [6]

    1. (a) (a) \(\int\frac{5}{3-2x}dx\) [2]

      • M1 for \(\ln|3-2x|\) with a constant factor
      • A1 for \(- \frac{5}{2}\ln|3-2x|+c\)

    2. (b) (b) \(\int\sin(4-3x)dx\) [2]

      • M1 for \(\cos\) of the same argument, divided by \(3\) (the sign needs care)
      • A1 for \(\frac{1}{3}\cos(4-3x)+c\)

    3. (c) (c) \(\int\frac{6x^{2}}{x^{3}+4}dx\) [2]

      • M1 for noticing that the numerator is a multiple of the derivative of the denominator (\(g'(x)=3x^{2}\))
      • A1 for \(2\ln\left(x^{3}+4\right)+c\)

  3. Find the exact value of each integral. Show your working. [6]

    1. (a) (a) \(\int_{0}^{\ln 3}4e^{2x}dx\) [3]

      • A1 for \(2e^{2x}\)
      • M1 for substituting the limits, using \(e^{2\ln 3}=9\)
      • A1 for \(16\)

    2. (b) (b) \(\int_{-4}^{-1}\frac{6}{x}dx\) [3]

      • A1 for \(6\ln|x|\)
      • M1 for \(6\ln|-1|-6\ln|-4|\)
      • A1 for \(-6\ln 4\) = \(-8.32\)

  4. The curve \(C\) has equation \(y=x^{3}-7x^{2}+10x\). [5]

    1. (a) (a) Write down the \(x\)-intercepts of \(C\). [1]

      • A1 for \(0,\ 2,\ 5\) (from the factorised form or the graph)

    2. (b) (b) Write down an expression for the total area of the regions enclosed by \(C\) and the \(x\)-axis. [2]

      • A1 for \(\int_{0}^{2}y\,dx\) (above the axis)
      • A1 for \(-\int_{2}^{5}y\,dx\) (the region below the axis is counted positive)

    3. (c) (c) Use your calculator to find the total area. [2]

      • M1 for evaluating both integrals (\(5.33\) and \(-15.75\)) or \(\int|y|dx\)
      • A1 for \(21.1\)

  5. Use the substitution given in each part. [6]

    1. (a) (a) Use the substitution \(u=x^{2}+3\) to show that \(\int_{0}^{1}x\sqrt{x^{2}+3}\,dx=\frac{8-3\sqrt{3}}{3}\). [4]

      • M1 for \(\frac{du}{dx}=2x\), so \(x\,dx=\frac{1}{2}du\)
      • A1 for the new limits \(u=3\) and \(u=4\)
      • M1 for \(\frac{1}{2}\int_{3}^{4}u^{\frac{1}{2}}du=\frac{1}{3}\left[u^{\frac{3}{2}}\right]_{3}^{4}\)
      • A1 for \(\frac{8-3\sqrt{3}}{3}\) (AG)

    2. (b) (b) Use the substitution \(u=1+\sin x\) to find \(\int\frac{\cos x}{1+\sin x}\,dx\). [2]

      • M1 for \(du=\cos x\,dx\) so the integral becomes \(\int\frac{1}{u}du\)
      • A1 for \(\ln|1+\sin x|+c\)

  6. The function \(f\) satisfies \(\int_{1}^{5}f(x)\,dx=12\). [6]

    1. (a) (a) Find \(\int_{1}^{5}\left(4f(x)+3\right)dx\). [2]

      • M1 for \(4\times 12+3\times 4\)
      • A1 for \(60\)

    2. (b) (b) Find \(\int_{2}^{10}f\left(\frac{x}{2}\right)dx\). [2]

      • M1 for a horizontal stretch of scale factor \(2\): the area is multiplied by \(2\)
      • A1 for \(2\times 12=24\)

    3. (c) (c) Given that \(\int_{1}^{5}\left(af(x)-2\right)dx=28\), find the value of \(a\). [2]

      • M1 for \(12a-2\times 4=28\)
      • A1 for \(a=3\)

  7. Find the exact value of each integral. Show your working. [6]

    1. (a) (a) \(\int_0^{\ln 3}6e^{-2x}dx\) [2]

      • A1 for \(-3e^{-2x}\)
      • A1 for \(\frac{8}{3}\) using \(e^{-2\ln 3}=\frac{1}{9}\)

    2. (b) (b) \(\int_{-4}^{-1}\frac{6}{x}dx\) [2]

      • A1 for \(6\ln|x|\)
      • A1 for \(-12\ln 2\) = -8.32

    3. (c) (c) \(\int_{0}^{3}\frac{6}{x-5}dx\) [2]

      • A1 for \(6\ln|x-5|\), with the modulus because \(x-5\) is negative on the interval
      • A1 for \(-6\ln 5 + 6 \ln 2\) = -5.5

  8. A function \(f\) satisfies \(f'(x)=\frac{6}{x}\) for \(x<0\) and \(f(-1)=2\). [6]

    1. (a) (a) Find the exact value of \(f(-3)\). [3]

      • A1 for \(f(x)=6\ln|x|+C\)
      • M1 for using \(f(-1)=2\) to find \(C=2\)
      • A1 for \(2 + 6 \ln 3\)

    2. (b) (b) Given that \(a>0\) and \(\int_a^{2a}(x+1)dx=8\), find the value of \(a\). [3]

      • M1 for \(\left[\frac{x^2}{2}+x\right]_a^{2a}\), giving \(\frac{3a^2}{2}+a=8\)
      • M1 for solving the quadratic \(3 a^{2} + 2 a - 16=0\)
      • A1 for \(a=2\), rejecting the negative root as \(a>0\)

  9. The graph of \(y=f(x)\) passes through the point \(P(4,-2)\). The function \(g\) is defined by \(g(x)=2f\left(\frac{x}{3}\right)+1\). It is given that \(\int_{0}^{4}f(x)\,dx=5\). [5]

    1. (a) (a) Describe the sequence of transformations that maps the graph of \(y=f(x)\) to the graph of \(y=g(x)\), and write down the coordinates of the image of \(P\). [2]

      • A1 for the three transformations: horizontal stretch scale factor 3, vertical stretch scale factor 2, translation of 1 in the \(y\)-direction
      • A1 for \((12,-3)\)

    2. (b) (b) Find \(\int_{0}^{12}g(x)\,dx\). [3]

      • M1 for the stretches multiplying the area by \(3\times 2=6\): \(6\times 5=30\)
      • M1 for the translation adding \(1\times 12=12\) (width \(12\))
      • A1 for \(42\)

Section B: long questions

  1. The diagram shows part of the curve \(C\) with equation \(y=\frac{6}{x+2}\), for \(0\leq x\leq 4\). The region \(R\) is bounded by \(C\), the \(x\)-axis, the \(y\)-axis and the line \(x=4\). The region \(S\) is bounded by \(C\), the \(y\)-axis and the lines \(y=1\) and \(y=3\). [14]

    123451234xyy = 6/(x + 2)O

    1. (a) (a) Find the area of \(R\). [3]

      • A1 for \(\int_{0}^{4}\frac{6}{x+2}dx=6\ln|x+2|\)
      • M1 for \(6\ln 6-6\ln 2\)
      • A1 for \(6\ln 3\) = \(6.59\)

    2. (b) (b) The region \(R\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume of the solid formed. [3]

      • M1 for \(\pi\int_{0}^{4}\frac{36}{(x+2)^{2}}dx\)
      • A1 for \(-\frac{36}{x+2}\)
      • A1 for \(12\pi\) = \(37.7\)

    3. (c) (c) Find the area of \(S\). [4]

      • A1 for \(x=\frac{6}{y}-2\)
      • M1 for \(\int_{1}^{3}x\,dy\)
      • A1 for \(6\ln y-2y\)
      • A1 for \(6\ln 3-4\) = \(2.59\)

    4. (d) (d) The region \(S\) is rotated through \(360^{\circ}\) about the \(y\)-axis. Find the volume of the solid formed. [4]

      • M1 for \(\pi\int_{1}^{3}x^{2}\,dy\)
      • A1 for \(\frac{36}{y^{2}}-\frac{24}{y}+4\)
      • A1 for \(-\frac{36}{y}-24\ln y+4y\)
      • A1 for \(\pi\left(32-24\ln 3\right)\) = \(17.7\)

  2. The diagram shows the graph of \(y=f(x)\), where \(f(x)=x^{3}-6x^{2}+8x\). The graph crosses the \(x\)-axis at \(x=0\), \(x=2\) and \(x=4\). The shaded region \(A\) lies between \(x=0\) and \(x=2\) and the shaded region \(B\) lies between \(x=2\) and \(x=4\). [12]

    1234−4−224xyy = f(x)24

    1. (a) (a) Find \(\int_{0}^{2}f(x)\,dx\). [2]

      • M1 for \(\left[\frac{x^{4}}{4}-6\frac{x^{3}}{3}+8\frac{x^{2}}{2}\right]_{0}^{2}\)
      • A1 for \(4\)

    2. (b) (b) Find the total area of the regions \(A\) and \(B\). [3]

      • M1 for evaluating \(\int_{2}^{4}f(x)\,dx\)
      • A1 for \(-4\) (below the axis)
      • A1 for \(4+4=8\)

    3. (c) (c) The function \(g\) is defined by \(g(x)=3f(x)+2\). Find \(\int_{0}^{2}g(x)\,dx\). [2]

      • M1 for \(3\times 4\) plus \(2\times 2\)
      • A1 for \(16\)

    4. (d) (d) The function \(h\) is defined by \(h(x)=f\left(\frac{x}{2}\right)\). Find \(\int_{0}^{4}h(x)\,dx\). [2]

      • M1 for a horizontal stretch of scale factor \(2\), so the area is multiplied by \(2\)
      • A1 for \(2\times 4=8\)

    5. (e) (e) Given that \(\int_{0}^{2}\left(kf(x)-3\right)dx=10\), find the value of \(k\). [2]

      • M1 for \(4k-3\times 2=10\)
      • A1 for \(k=4\)

    6. (f) (f) Find \(\int_{3}^{5}f(x-3)\,dx\). [1]

      • A1 for a translation of \(3\) to the right does not change the area: \(4\)

  3. The region \(R\) is enclosed by the parabola \(y=x^{2}\) and the line \(y=2x\). Units are centimetres. [13]

    −0.50.511.522.512345xyy = x²y = 2xO

    1. (a) (a) Find the coordinates of the points where the line and the parabola meet. [2]

      • M1 for \(x^{2}=2x\)
      • A1 for \((0,0)\) and \((2,4)\)

    2. (b) (b) Find the area of \(R\). [2]

      • M1 for \(\int_{0}^{2}(2x-x^{2})dx\)
      • A1 for \(\frac{4}{3}\) = \(1.33\)

    3. (c) (c) \(R\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the volume of the solid formed. [3]

      • M1 for \(\pi\int_{0}^{2}\left((2x)^{2}-(x^{2})^{2}\right)dx\)
      • A1 for \(\pi\left[\frac{4}{3}x^{3}-\frac{x^{5}}{5}\right]_{0}^{2}\)
      • A1 for \(\frac{64 \pi}{15}\) = \(13.4\)

    4. (d) (d) \(R\) is rotated through \(360^{\circ}\) about the \(y\)-axis. Find the volume of the solid formed. [4]

      • A1 for \(x=\sqrt{y}\) (parabola) and \(x=\frac{y}{2}\) (line)
      • M1 for \(\pi\int_{0}^{4}\left(y-\frac{y^{2}}{4}\right)dy\)
      • A1 for \(\pi\left[\frac{y^{2}}{2}-\frac{y^{3}}{12}\right]_{0}^{4}\)
      • A1 for \(\frac{8 \pi}{3}\) = \(8.38\)

    5. (e) (e) The part of the line \(y=2x\) from \(x=0\) to \(x=a\), \(a>0\), is rotated through \(360^{\circ}\) about the \(x\)-axis. The volume of the cone formed is \(36\pi\). Find the value of \(a\). [2]

      • M1 for \(\pi\int_{0}^{a}(2x)^{2}dx=\frac{4}{3}\pi a^{3}\)
      • A1 for \(\frac{4}{3}a^{3}=36\) so \(a=3\)

  4. The concentration, \(C\) mg per litre, of a drug in a patient's blood \(t\) hours after an injection satisfies \(\frac{dC}{dt}=\frac{6t}{t^{2}+4}\), \(t\geq 0\). Initially the concentration is 2 mg per litre. [13]

    1. (a) (a) Use the substitution \(u=t^{2}+4\) to find \(C(t)\). [4]

      • M1 for \(du=2t\,dt\) so \(\int\frac{6t}{t^{2}+4}dt=3\int\frac{1}{u}du\)
      • A1 for \(3\ln(t^{2}+4)+c\)
      • M1 for using \(C(0)=2\), so \(c=2-3\ln 4\)
      • A1 for \(C(t)=3\ln\left(\frac{t^{2}+4}{4}\right)+2\) (or an equivalent form)

    2. (b) (b) Find the value of \(C\) when \(t=6\). [2]

      • M1 for substituting \(t=6\)
      • A1 for \(8.91\)

    3. (c) (c) Find the exact change in \(C\) between \(t=2\) and \(t=6\). [3]

      • M1 for \(\int_{2}^{6}\frac{6t}{t^{2}+4}dt\) or \(C(6)-C(2)\)
      • A1 for \(3\left(\ln 40-\ln 8\right)\)
      • A1 for \(3\ln 5\) = \(4.83\)

    4. (d) (d) Use your calculator to find the value of \(t\) for which \(C=10\). [2]

      • M1 for solving \(C(t)=10\) (graph intersection or solver)
      • A1 for \(t=7.32\)

    5. (e) (e) Find the value that \(\frac{dC}{dt}\) approaches as \(t\to\infty\). Comment on the long-term behaviour of \(C\). [2]

      • A1 for \(\frac{dC}{dt}\to 0\)
      • R1 for \(C\) keeps increasing because the rate is always positive and \(\ln(t^{2}+4)\) has no upper limit

  5. A glass studio designs a vase. Units are decimetres, so that 1 dm\(^3\) is one litre. A vase is made by rotating the region bounded by the curve \(y=\sqrt{x^2+1}\), the \(y\)-axis and the line \(y=3\) through \(360^{\circ}\) about the \(y\)-axis. [11]

    1. (a) (a) Show that \(x^2=y^2-1\). [1]

      • R1 for squaring both sides: \(y^2=x^2+1\) so \(x^2=y^2-1\)

    2. (b) (b) Find the exact capacity of the vase, in dm\(^3\). [3]

      • M1 for \(\pi\int_{1}^{3}(y^2-1)dy\)
      • A1 for \(\pi\left[\frac{y^3}{3}-y\right]_{1}^{3}\)
      • A1 for \(\frac{20 \pi}{3}\)

    3. (c) (c) Write down the capacity of the vase in litres, correct to 3 significant figures. [1]

      • A1 for 20.9 litres

    4. (d) (d) The vase is filled with 10 litres of water. Find the depth of the water, to 3 significant figures. [3]

      • M1 for \(\pi\int_{1}^{h}(y^2-1)dy=10\)
      • M1 for solving the equation with technology
      • A1 for \(h=2.46\)

    5. (e) (e) Find the area of the region used to make the vase, giving your answer to 3 significant figures. [3]

      • M1 for \(\int_{1}^{3}\sqrt{y^2-1}\,dy\) (x as a function of y)
      • (A1) for the correct limits
      • A1 for 3.36

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