Further calculus – further maths worksheet answers to download as PDF

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1. Answer to Question 1

   Method: an improper integral has an infinite limit or an integrand that is unbounded at an end point. Replace the problem limit by a letter (\(R\) or \(a\)), integrate normally, then let the letter tend to its limit. If the limit is finite, the integral converges.

   1. Replace \(\infty\) by \(R\): \(\left[-\tfrac1x\right]_1^R\). \(\int_1^R x^{-2}dx=1-\dfrac1R\to\) 1 as \(R\to\infty\). Converges. So \(\int_1^\infty\dfrac{dx}{x^2}=1\).
   2. The integral of \(\tfrac1x\) is \(\ln x\), which has no upper bound. \(\int_1^R\dfrac1x\,dx=\ln R\to\infty\). Diverges. Although \(\tfrac1x\to0\), it does not tend to 0 fast enough.
   3. The integrand is unbounded at 0, so replace 0 by \(a\) and let \(a\to0^+\). \(\int_a^1x^{-1/2}dx=2-2\sqrt a\to\) 2 as \(a\to0^+\). Converges. So the area under \(\tfrac1{\sqrt x}\) from 0 to 1 is finite and equal to 2.
   4. Replace \(\infty\) by \(R\); \(e^{-3R}\to0\). \(\int_0^Re^{-3x}dx=\dfrac13\left(1-e^{-3R}\right)\to\) \(\dfrac13\). Converges. So the integral equals \(\tfrac13\).

2. Answer to Question 2

   1. By parts: \(\int xe^{-x}dx=-xe^{-x}-e^{-x}\). Between 0 and \(R\): \(1-(R+1)e^{-R}\to\) 1.
   2. \(\int\ln x\,dx=x\ln x-x\). From \(a\) to 1: \(-1-a\ln a+a\to\) \(-1\) since \(a\ln a\to0\).
   3. The integrand is undefined at \(x=1\), inside the interval. Split: \(\int_0^{1-}\dfrac1{(x-1)^2}dx=\left[\dfrac{-1}{x-1}\right]\) grows without limit. Diverges (so the answer \(-2\) from the naive calculation is wrong).
   4. \(\dfrac1{x(x+1)}=\dfrac1x-\dfrac1{x+1}\). \(\left[\ln\dfrac x{x+1}\right]_1^R=\ln\dfrac R{R+1}+\ln2\to\) \(\ln2\).

3. Answer to Question 3

   Method: use \(\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+c\) and \(\displaystyle\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+c\). If the quadratic is not in this form, complete the square first and then substitute \(u=x+k\).

   1. Here \(a=3\): \(\arcsin\dfrac x3+C\).
   2. Take out the factor 4 so that the denominator has the form \(x^2+a^2\) with \(a=\tfrac12\). \(\dfrac1{1+4x^2}=\dfrac14\cdot\dfrac1{x^2+(1/2)^2}\). The integral is \(\dfrac14\cdot2\arctan2x=\tfrac12\arctan2x\). Between 0 and 1: \(\tfrac12\arctan2\approx0.5536\).
   3. Here \(a=2\). \(\left[\arcsin\dfrac x2\right]_0^{\sqrt2}=\arcsin\dfrac{\sqrt2}2=\) \(\dfrac\pi4\).
   4. \(x^2+2x+5=(x+1)^2+4\). \(\left[\tfrac12\arctan\dfrac{x+1}2\right]_{-1}^{1}=\tfrac12\arctan1-0=\) \(\dfrac\pi8\). So the integral equals \(\tfrac\pi8\approx0.393\).

4. Answer to Question 4

   Method: rotating about the \(x\)-axis gives \(V=\pi\displaystyle\int_a^by^2\,dx\); about the \(y\)-axis, \(V=\pi\displaystyle\int_c^dx^2\,dy\), with the limits in \(y\). Square the function first, then integrate.

   1. Square \(y=\sqrt x\) to get \(y^2=x\). \(V=\pi\int_0^4x\,dx=\pi\left[\tfrac{x^2}2\right]_0^4=\) \(8\pi\) So the volume is \(8\pi\approx25.1\) cubic units.
   2. Square \(y=x^2\) to get \(y^2=x^4\). \(V=\pi\int_0^2x^4dx=\pi\cdot\dfrac{32}5=\) \(\dfrac{32\pi}5\) So the volume is \(\tfrac{32\pi}5\approx20.1\) cubic units.
   3. Rotating about the \(y\)-axis, write \(x^2\) in terms of \(y\) and use the limits 0 and 4 for \(y\). \(x^2=y\), so \(V=\pi\int_0^4y\,dy=\pi\cdot8=\) \(8\pi\) So the volume is \(8\pi\approx25.1\) cubic units.
   4. Square \(y=e^x\) to get \(e^{2x}\). \(V=\pi\int_0^1e^{2x}dx=\pi\cdot\dfrac{e^2-1}2=\) \(\dfrac{\pi(e^2-1)}2\) So the volume is about 10.0 cubic units.

5. Answer to Question 5

   1. \(x=x^2\) gives \(x=0\) or \(x=1\); the line \(y=x\) lies above the parabola between them.
   2. Rotating the line alone gives a cone of cross-section \(\pi x^2\); rotating the curve alone gives a solid with cross-section \(\pi(x^2)^2=\pi x^4\). The required solid is the cone with the second solid removed, so the cross-section is a ring with area \(\pi x^2-\pi x^4\).
   3. \(V=\pi\left[\dfrac{x^3}3-\dfrac{x^5}5\right]_0^1=\pi\left(\dfrac13-\dfrac15\right)=\) \(\dfrac{2\pi}{15}\)

6. Answer to Question 6

   Method: the mean value of \(f\) on \([a,b]\) is \(\dfrac1{b-a}\displaystyle\int_a^bf(x)\,dx\): the height of the rectangle on \([a,b]\) with the same area as the region under the curve.

   1. \(\sin^2x=\tfrac12(1-\cos2x)\), so \(\int_0^\pi=\tfrac\pi2\) and the mean is \(\dfrac1\pi\cdot\dfrac\pi2=\) \(\dfrac12\)
   2. \(\int_1^3x^2dx=\dfrac{27-1}3=\dfrac{26}3\), mean \(=\dfrac12\cdot\dfrac{26}3=\) \(\dfrac{13}3\)
   3. \(\int_0^2e^{-x}dx=1-e^{-2}\), mean \(=\dfrac{1-e^{-2}}2\approx\) 0.4323
   4. \(\int_0^1\dfrac{dx}{1+x^2}=\dfrac\pi4\) and \(b-a=1\), so the mean is \(\dfrac\pi4\approx0.785\)

7. Answer to Question 7

   1. \(\displaystyle M(T)=\int_0^T40e^{-0.5t}dt\).
   2. \(M(T)=80\left(1-e^{-0.5T}\right)\). At \(T=2\): \(80(1-e^{-1})=\) 50.6 mg.
   3. The absorption carries on for ever (it never exactly stops), so we take \(\lim_{T\to\infty}M(T)=\displaystyle\int_0^\infty40e^{-0.5t}dt=\) 80 mg, the mass of the tablet.
   4. \(\dfrac{50.57}{80}=\) 63.2%, which is \(1-e^{-1}\).

8. Answer to Question 8

   Method: rotating about the \(x\)-axis gives \(V=\pi\displaystyle\int_a^by^2\,dx\); about the \(y\)-axis, \(V=\pi\displaystyle\int_c^dx^2\,dy\), with the limits in \(y\). Square the function first, then integrate.

   1. \(V=\pi\int_0^hx^2\,dy=\pi\int_0^hy\,dy=\dfrac{\pi h^2}2\).
   2. At \(h=8\): \(\dfrac{64\pi}2=32\pi\approx\) 101 cm\(^3\).
   3. The radius is \(x=\sqrt y=\sqrt6\approx\) 2.45 cm.
   4. Volume at 6 cm: \(18\pi=56.55\) cm\(^3\). Time \(=\dfrac{56.55}{20}\approx\) 2.83 s.

9. Answer to Question 9

   Method: rotating about the \(x\)-axis gives \(V=\pi\displaystyle\int_a^by^2\,dx\); about the \(y\)-axis, \(V=\pi\displaystyle\int_c^dx^2\,dy\), with the limits in \(y\). Square the function first, then integrate.

   1. Substitute \(x=9\) into the equation of the line. At \(x=9\), \(y=4\): radius 4 cm, height 9 cm.
   2. Square \(y\) and integrate from 0 to 9. \(V=\pi\int_0^9\left(\tfrac49x\right)^2dx=\pi\cdot\tfrac{16}{81}\cdot\dfrac{9^3}3=\pi\cdot\tfrac{16}{81}\cdot243=\) \(48\pi\) cm\(^3\) (\(\approx150.8\)).
   3. Use the radius and height from part a). \(\tfrac13\pi\cdot16\cdot9=48\pi\). ✓ The two methods agree.
   4. Mass is density times volume, with density \(1.2\) g per cm³. Mass \(=1.2\times150.8=\) 181 g. So the cone holds about 181 g of ice cream.

10. Answer to Question 10

    Method: rotating about the \(x\)-axis gives \(V=\pi\displaystyle\int_a^by^2\,dx\); about the \(y\)-axis, \(V=\pi\displaystyle\int_c^dx^2\,dy\), with the limits in \(y\). Square the function first, then integrate.

    1. Square both sides of \(y=\sqrt{100-x^2}\). \(x^2+y^2=100\) with \(y\ge0\): a semicircle with centre the origin and radius 10 m.
    2. Square \(y\) and integrate from 0 to 10. \(V=\pi\int_0^{10}(100-x^2)dx=\pi\left[100x-\dfrac{x^3}3\right]_0^{10}=\pi\left(1000-\dfrac{1000}3\right)=\) \(\dfrac{2000\pi}3\approx2094\) m\(^3\).
    3. \(\tfrac23\pi\cdot10^3=\dfrac{2000\pi}3\). ✓ The two methods agree.
    4. Multiply the volume by the power needed per cubic metre. \(0.08\times2094\approx\) 168 kW. So about 168 kW of heating is needed.

11. Answer to Question 11

    1. \(t=0\): 14\(^\circ\)C. \(t=6\): \(14+6\sin\tfrac\pi2=\) 20\(^\circ\)C. \(t=12\): \(14+6\sin\pi=\) 14\(^\circ\)C.
    2. \(\bar T=\dfrac1{12}\displaystyle\int_0^{12}\left(14+6\sin\dfrac{\pi t}{12}\right)dt\).
    3. \(\int_0^{12}6\sin\dfrac{\pi t}{12}dt=6\cdot\dfrac{12}\pi\cdot2=\dfrac{144}\pi\). So \(\bar T=14+\dfrac{12}\pi=\) 17.8\(^\circ\)C.
    4. The temperature is at least 14 and reaches a maximum of 20, so the mean must lie between 14 and 20; since it spends time above 14, it exceeds 14.

12. Answer to Question 12

    Method: the mean value of \(f\) on \([a,b]\) is \(\dfrac1{b-a}\displaystyle\int_a^bf(x)\,dx\): the height of the rectangle on \([a,b]\) with the same area as the region under the curve.

    1. \(\sin^2A=\tfrac12(1-\cos2A)\), so \(P=50(1-\cos240\pi t)\).
    2. The period of \(\cos240\pi t\) is \(\dfrac{2\pi}{240\pi}=\dfrac1{120}\) s, so \(\tfrac1{120}\) s (0.00833 s).
    3. \(\bar P=120\displaystyle\int_0^{1/120}\left(50-50\cos240\pi t\right)dt=120\left[50t-\dfrac{50\sin240\pi t}{240\pi}\right]_0^{1/120}=120\cdot\dfrac{50}{120}=\) 50 W.
    4. Energy \(=50\times60=\) 3000 J.

13. Answer to Question 13

    Method: an improper integral has an infinite limit or an integrand that is unbounded at an end point. Replace the problem limit by a letter (\(R\) or \(a\)), integrate normally, then let the letter tend to its limit. If the limit is finite, the integral converges.

    1. Integrate the current from 0 to 1. \(Q=\displaystyle\int_0^13e^{-2t}dt=\dfrac32\left(1-e^{-2}\right)\approx\) 1.30 C.
    2. Let the upper limit tend to infinity: \(e^{-2R}\to0\). \(\displaystyle\int_0^\infty3e^{-2t}dt=\) 1.5 C. So in total 1.5 coulombs flow.
    3. Set the charge equal to half of 1.5 C and solve with logarithms. \(\dfrac32\left(1-e^{-2T}\right)=0.75\), \(e^{-2T}=\tfrac12\), \(T=\tfrac12\ln2\approx\) 0.347 s. So half the charge has flowed after about 0.35 s.
    4. Use the mean value formula on \([0,1]\). Mean \(=\dfrac{1.297}{1}=\) 1.30 A (the mean over one second equals the charge in that second).

14. Answer to Question 14

    Method: an improper integral has an infinite limit or an integrand that is unbounded at an end point. Replace the problem limit by a letter (\(R\) or \(a\)), integrate normally, then let the letter tend to its limit. If the limit is finite, the integral converges.

    1. \(\dfrac1\pi\left[\arctan x\right]_{-\infty}^\infty=\dfrac1\pi\left(\dfrac\pi2+\dfrac\pi2\right)=\) 1.
    2. \(\dfrac1\pi\left[\arctan x\right]_0^1=\dfrac1\pi\cdot\dfrac\pi4=\) \(\dfrac14\).
    3. \(\dfrac1\pi\left(\dfrac\pi2-\dfrac\pi4\right)=\) \(\dfrac14\).
    4. \(\int_0^R\dfrac x{\pi(1+x^2)}dx=\dfrac1{2\pi}\ln(1+R^2)\to\infty\), so the integral on each side diverges. The mean is undefined even though the distribution is a valid one.

15. Answer to Question 15

    1. For \(p\ne1\), \(\int_1^Rx^{-p}dx=\dfrac{R^{1-p}-1}{1-p}\). If \(p>1\) then \(R^{1-p}\to0\) and the limit is \(\dfrac1{p-1}\). If \(p<1\) then \(R^{1-p}\to\infty\). If \(p=1\) the integral is \(\ln R\to\infty\). Converges only for \(p>1\).
    2. \(\dfrac1{p-1}=\dfrac12\), so \(p=3\).
    3. \(\int_a^1x^{-p}dx=\dfrac{1-a^{1-p}}{1-p}\) for \(p\ne1\). If \(p<1\) then \(a^{1-p}\to0\) and the limit is \(\dfrac1{1-p}\). If \(p\ge1\) the limit does not exist (for \(p=1\) it is \(-\ln a\to\infty\)). Converges only for \(p<1\).
    4. \(x=u^2\), \(dx=2u\,du\), \(\sqrt x=u\): \(\displaystyle\int_0^\infty\dfrac{2u\,du}{u(1+u^2)}=2\left[\arctan u\right]_0^\infty=\) \(\pi\).

16. Answer to Question 16

    Method: rotating about the \(x\)-axis gives \(V=\pi\displaystyle\int_a^by^2\,dx\); about the \(y\)-axis, \(V=\pi\displaystyle\int_c^dx^2\,dy\), with the limits in \(y\). Square the function first, then integrate. For the region between two curves, use \(\pi\displaystyle\int(y_1^2-y_2^2)\,dx\), where \(y_1\) is the outer curve.

    1. Solve the two equations simultaneously. \(x^2=2x\) gives \(x=0\) or \(2\): \((0,0)\) and \((2,4)\).
    2. The line is above the curve between the two points, so integrate line minus curve. \(\displaystyle\int_0^2(2x-x^2)dx=\left[x^2-\dfrac{x^3}3\right]_0^2=4-\dfrac83=\) \(\dfrac43\). So the area of \(R\) is \(\tfrac43\).
    3. The outer radius is \(2x\) and the inner radius is \(x^2\). Do not square the difference. \(V=\pi\displaystyle\int_0^2\left((2x)^2-(x^2)^2\right)dx=\pi\left[\dfrac{4x^3}3-\dfrac{x^5}5\right]_0^2=\pi\left(\dfrac{32}3-\dfrac{32}5\right)=\) \(\dfrac{64\pi}{15}\). So the volume is \(\tfrac{64\pi}{15}\approx13.4\).
    4. 1 dm³ is 1 litre. \(\dfrac{64\pi}{15}\approx\) 13.4 litres.

17. Answer to Question 17

    1. The curve starts at \(y=1\), rises to \(y=2\) at \(x=\tfrac\pi2\) and returns to \(y=1\) at \(x=\pi\). The vase is widest in the middle (radius 2 cm) and has radius 1 cm at both ends, like a barrel.
    2. \(y^2=1+2\sin x+\sin^2x=1+2\sin x+\tfrac12(1-\cos2x)=\tfrac32+2\sin x-\tfrac12\cos2x\).
    3. \(V=\pi\left[\tfrac32x-2\cos x-\tfrac14\sin2x\right]_0^\pi=\pi\left(\tfrac{3\pi}2+2-(-2)\right)=\) \(\pi\left(4+\tfrac{3\pi}2\right)\).
    4. \(V=\pi(4+4.712)\approx\) 27.4 cm\(^3\). A cup holds 250 cm\(^3\), so the vase fills 0.11 of a cup, far less than one cup: the vase is small.

18. Answer to Question 18

    1. \(I_0=\int_0^\infty e^{-x}dx=\left[-e^{-x}\right]_0^\infty=\) 1.
    2. Let \(u=x^n\), \(dv=e^{-x}dx\): \(I_n=\left[-x^ne^{-x}\right]_0^\infty+n\displaystyle\int_0^\infty x^{n-1}e^{-x}dx\). The boundary term is 0 at \(x=0\) (for \(n\ge1\)) and tends to 0 as \(x\to\infty\) because exponentials grow faster than any power. Hence \(I_n=nI_{n-1}\).
    3. \(I_1=1\cdot1=\) 1, \(I_2=2\cdot1=\) 2, \(I_3=3\cdot2=\) 6, \(I_4=4\cdot6=\) 24.
    4. \(I_n=n!\). (This is the idea behind the gamma function.)

19. Answer to Question 19

    Method: use \(\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+c\) and \(\displaystyle\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+c\). If the quadratic is not in this form, complete the square first and then substitute \(u=x+k\).

    1. \(4x-x^2=4-(x-2)^2\), so the integral is \(\left[\arcsin\dfrac{x-2}2\right]_0^4=\arcsin1-\arcsin(-1)=\dfrac\pi2+\dfrac\pi2=\) \(\pi\). (It is improper because the integrand is unbounded at both ends, yet converges.)
    2. \(x^2-4x+13=(x-2)^2+9\). \(\left[\tfrac13\arctan\dfrac{x-2}3\right]_2^\infty=\tfrac13\cdot\dfrac\pi2=\) \(\dfrac\pi6\).
    3. \(\displaystyle\int_2^5\dfrac{dx}{(x-2)^2+9}=\tfrac13\arctan1=\dfrac\pi{12}\). Interval length 3. Mean \(=\) \(\dfrac\pi{36}\approx0.0873\).

20. Answer to Question 20

    1. \(V=\pi\displaystyle\int_1^\infty\dfrac1{x^2}dx=\pi\cdot1=\) \(\pi\).
    2. \(\dfrac{dy}{dx}=-\dfrac1{x^2}\), so \(\sqrt{1+\dfrac1{x^4}}>1\) and \(y\sqrt{\dots}>\dfrac1x\) for \(x\ge1\). Hence \(A>2\pi\int_1^\infty\dfrac1x\,dx\).
    3. \(\int_1^\infty\dfrac1x\,dx\) diverges (it is \(\ln R\to\infty\)), so by comparison \(A\) diverges.
    4. A finite volume of paint spreads over a surface of infinite area if the layer of paint gets thinner and thinner. A real layer cannot become thinner than one molecule, so the paradox disappears in the physical world; it is a statement about idealised mathematics only.