Integration and the fundamental theorem of calculus – university Year 1 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Integrate term by term with the power rule \(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}\): \(\int3x^2\,dx=x^3\), \(\int2x\,dx=x^2\) and \(\int1\,dx=x\). So the answer is \(x^3-x^2+x+C\). Differentiating gives back \(3x^2-2x+1\), which checks it.
   2. An antiderivative is \(\dfrac{x^3}3+x\). By the fundamental theorem, \(\left[\dfrac{x^3}3+x\right]_0^2=\left(\dfrac83+2\right)-0=\) \(\dfrac{14}3\).
   3. Write \(\sqrt x=x^{1/2}\), so an antiderivative is \(\dfrac23x^{3/2}\). Then \(\left[\dfrac23x^{3/2}\right]_1^4=\dfrac23(8-1)=\) \(\dfrac{14}3\), since \(4^{3/2}=8\).
   4. Since \(\dfrac d{dx}e^{2x}=2e^{2x}\), we have \(\int e^{2x}dx=\dfrac12e^{2x}\); and \(\int\dfrac1x\,dx=\ln x\) for \(x>0\). The answer is \(\dfrac12e^{2x}+\ln x+C\).

2. Answer to Question 2

   1. The numerator is the derivative of the denominator, so let \(u=x^2+1\), with \(du=2x\,dx\). The integral becomes \(\int\dfrac{du}u=\ln|u|+C\). Since \(x^2+1>0\), the answer is \(\ln(x^2+1)+C\).
   2. Let \(u=x^2\), so \(du=2x\,dx\) and \(x\,dx=\dfrac12du\). Change the limits: \(x=0\) gives \(u=0\) and \(x=1\) gives \(u=1\). Then \(\dfrac12\int_0^1e^u\,du=\dfrac12\left[e^u\right]_0^1=\dfrac12(e-1)=\) \(\dfrac{e-1}2\), about \(0.859\).
   3. Let \(u=\sin x\), so \(du=\cos x\,dx\). The integral becomes \(\int u^3du=\dfrac{u^4}4+C\). Substituting back gives \(\dfrac{\sin^4x}4+C\). Differentiating with the chain rule returns \(\sin^3x\cos x\).

3. Answer to Question 3

   1. Take \(u=x\), \(v'=e^x\). Then \(xe^x-\int e^xdx=\) \((x-1)e^x+C\).
   2. Take \(u=\ln x\), \(v'=x^2\), \(v=\dfrac{x^3}3\): \(\dfrac{x^3}3\ln x-\int\dfrac{x^3}3\cdot\dfrac1x\,dx=\dfrac{x^3}3\ln x-\dfrac{x^3}9+C=\) \(\dfrac{x^3}9(3\ln x-1)+C\).
   3. Take \(u=x\), \(v'=\sin x\), \(v=-\cos x\): \(-x\cos x+\int\cos x\,dx=\) \(\sin x-x\cos x+C\).
   4. Write \(\ln x=1\cdot\ln x\) with \(u=\ln x\), \(v'=1\): \(x\ln x-\int1\,dx=\) \(x\ln x-x+C\).

4. Answer to Question 4

   1. Factorise \(x^2-1=(x-1)(x+1)\) and write \(\dfrac1{(x-1)(x+1)}=\dfrac A{x-1}+\dfrac B{x+1}\). Multiplying out, \(1=A(x+1)+B(x-1)\). Putting \(x=1\) gives \(A=\tfrac12\); putting \(x=-1\) gives \(B=-\tfrac12\). Integrating each term, \(\tfrac12\ln(x-1)-\tfrac12\ln(x+1)+C=\) \(\dfrac12\ln\dfrac{x-1}{x+1}+C\), using the laws of logarithms.
   2. Write \(\dfrac{3x+5}{(x+1)(x+2)}=\dfrac A{x+1}+\dfrac B{x+2}\), so \(3x+5=A(x+2)+B(x+1)\). At \(x=-1\): \(2=A\). At \(x=-2\): \(-1=-B\), so \(B=1\). Integrating, the answer is \(2\ln(x+1)+\ln(x+2)+C\), which may also be written \(\ln\left[(x+1)^2(x+2)\right]+C\).

5. Answer to Question 5

   1. There is no direct antiderivative of \(\sin^2x\), so use the double-angle identity \(\sin^2x=\dfrac{1-\cos2x}2\). Integrating each term, the answer is \(\dfrac x2-\dfrac{\sin2x}4+C\).
   2. Odd powers of sine are handled by keeping one factor: \(\sin^3x=\sin x(1-\cos^2x)=\sin x-\sin x\cos^2x\). With \(u=\cos x\), \(\int\sin x\cos^2x\,dx=-\dfrac{\cos^3x}3\), so an antiderivative is \(-\cos x+\dfrac{\cos^3x}3\). Then \(\left[-\cos x+\dfrac{\cos^3x}3\right]_0^\pi=\left(1-\dfrac13\right)-\left(-1+\dfrac13\right)=\) \(\dfrac43\).
   3. Use the standard result \(\int\dfrac{dx}{x^2+a^2}=\dfrac1a\arctan\dfrac xa+C\) with \(a=2\). The answer is \(\dfrac12\arctan\dfrac x2+C\).

6. Answer to Question 6

   1. Let \(F(u)=\int_0^ue^{t^2}dt\). By the fundamental theorem, \(F'(u)=e^{u^2}\). The function we differentiate is \(F(x^2)\), so by the chain rule \(\dfrac d{dx}F(x^2)=F'(x^2)\cdot2x=\) \(2xe^{x^4}\).
   2. The variable is in the lower limit, so swap the limits: \(\int_x^3\sin(t^2)\,dt=-\int_3^x\sin(t^2)\,dt\). By the fundamental theorem the derivative is \(-\sin(x^2)\).
   3. With 4 strips the width is \(\frac14\), and the left endpoints are \(0,\frac14,\frac12,\frac34\). The sum is \(\frac14\left(0+\frac1{16}+\frac14+\frac9{16}\right)=\frac14\cdot\frac{14}{16}=\frac7{32}=\) 0.21875. The exact value is \(\left[\frac{x^3}3\right]_0^1=\frac13\approx0.3333\). The left sum is an underestimate because \(x^2\) is increasing on \([0,1]\), so each rectangle lies below the curve.

7. Answer to Question 7

   1. \(\displaystyle\int_0^4(t^2-4t+3)dt=\left[\dfrac{t^3}3-2t^2+3t\right]_0^4=\dfrac{64}3-32+12=\) \(\dfrac43\) m.
   2. \(t^2-4t+3=(t-1)(t-3)=0\) at \(t=1\) and \(t=3\).
   3. Let \(F(t)=\dfrac{t^3}3-2t^2+3t\): \(F(0)=0\), \(F(1)=\dfrac43\), \(F(3)=0\), \(F(4)=\dfrac43\). The velocity is positive on \((0,1)\), negative on \((1,3)\), positive on \((3,4)\). Distance \(=\left|\dfrac43\right|+\left|-\dfrac43\right|+\dfrac43=\) 4 m.
   4. Displacement counts backward motion as negative and so cancels it; distance adds up all the movement regardless of direction.

8. Answer to Question 8

   1. The volume is the accumulated inflow: \(V(t)=\int_0^t12e^{-0.1s}ds\). An antiderivative of \(12e^{-0.1s}\) is \(-120e^{-0.1s}\), so \(V(t)=\left[-120e^{-0.1s}\right]_0^t=\) \(120\left(1-e^{-0.1t}\right)\) litres.
   2. Substituting \(t=30\): \(V(30)=120(1-e^{-3})\approx\) 114.0 litres.
   3. Solve \(120(1-e^{-0.1t})=100\): \(e^{-0.1t}=\dfrac16\), so \(-0.1t=-\ln6\) and \(t=10\ln6\approx\) 17.9 minutes.
   4. As \(t\to\infty\), \(e^{-0.1t}\to0\), so \(V\to\) 120 litres. This is the improper integral \(\int_0^\infty12e^{-0.1t}dt=120\): the inflow slows down so fast that the total stays finite.

9. Answer to Question 9

   1. \(\dfrac1{24}\int_0^{24}\left(12-6\cos\dfrac{\pi t}{12}\right)dt=\dfrac1{24}\left[12t-\dfrac{72}\pi\sin\dfrac{\pi t}{12}\right]_0^{24}=\dfrac{288}{24}=\) 12 \(^\circ\)C (the sine term vanishes over a full period).
   2. \(\dfrac1{12}\left[12t-\dfrac{72}\pi\sin\dfrac{\pi t}{12}\right]_6^{18}=\dfrac1{12}\left(144-\dfrac{72}\pi(-1-1)\right)=12+\dfrac{12}\pi\approx\) 15.8 \(^\circ\)C.
   3. Between 06:00 and 18:00 the cosine term is mostly negative (the temperature is above 12 around noon), so we exclude the cold night hours.
   4. \(\cos\dfrac{\pi t}{12}=0\) gives \(\dfrac{\pi t}{12}=\dfrac\pi2\) or \(\dfrac{3\pi}2\), so 06:00 and 18:00.

10. Answer to Question 10

    1. \(x=x^2\) gives \(x=0\) or \(x=1\): the points \((0,0)\) and \((1,1)\).
    2. Area \(=\int_0^1(x-x^2)dx=\left[\dfrac{x^2}2-\dfrac{x^3}3\right]_0^1=\dfrac16\) square units. One square unit is \(10\,000\) m\(^2\), so the area is \(\dfrac{10\,000}6\approx\) 1 667 m\(^2\).
    3. \(1667\times0.40=666.7\), so about £670 to the nearest £10.
    4. The road side is the diagonal of a unit square: \(\sqrt2\times100\approx\) 141 m.

11. Answer to Question 11

    1. Cut the vase into thin slices perpendicular to the \(x\)-axis. Each slice of thickness \(dx\) is a disc of radius \(y=\sqrt x\), so its volume is \(\pi y^2dx=\pi x\,dx\). Adding the slices, \(V=\pi\int_0^{16}x\,dx=\pi\left[\dfrac{x^2}2\right]_0^{16}=128\pi\approx\) 402 cm\(^3\).
    2. Since 1 cm\(^3\) = 1 ml, the volume is about 402 ml.
    3. Mass = density \(\times\) volume \(=402.1\times1.6\approx\) 643 g of sand.
    4. The open end is at \(x=16\), where the radius is \(y=\sqrt{16}=\) 4 cm, so the opening has diameter 8 cm.

12. Answer to Question 12

    1. \(\int_0^{0.3}40x\,dx=20(0.09)=\) 1.8 J.
    2. \(\int_{0.3}^{0.5}40x\,dx=20(0.25-0.09)=\) 3.2 J.
    3. The force grows with extension, so each extra centimetre costs more energy. The second stretch is shorter (0.2 m) but happens at larger forces (12 N to 20 N).
    4. \(20x^2=10\) gives \(x^2=0.5\), so \(x=\) 0.707 m.

13. Answer to Question 13

    1. Using the given result, \(\displaystyle\int_0^\infty f(x)\,dx=\frac14\int_0^\infty xe^{-x/2}dx=\frac14\times4=1\). Since also \(f(x)\ge0\), \(f\) is a valid probability density.
    2. Integrate by parts with \(u=x\) and \(v'=e^{-x/2}\), so \(v=-2e^{-x/2}\): \(\int_0^2xe^{-x/2}dx=\left[-2xe^{-x/2}\right]_0^2+2\int_0^2e^{-x/2}dx=-4e^{-1}+2\left[-2e^{-x/2}\right]_0^2=-4e^{-1}-4e^{-1}+4=4-8e^{-1}\). Dividing by 4: \(P(X\lt 2)=1-\dfrac2e\approx\) 0.264.
    3. The mean is \(\dfrac14\int_0^\infty x^2e^{-x/2}dx\). Using \(\int_0^\infty x^ne^{-ax}dx=\dfrac{n!}{a^{n+1}}\) with \(n=2\), \(a=\frac12\), this is \(\dfrac14\cdot\dfrac{2}{(1/2)^3}=\dfrac14\cdot16=\) 4 minutes.
    4. \(P(X\lt 2)\approx0.264\), so \(P(X>2)\approx\) 0.736. She is much more likely to wait more than 2 minutes.

14. Answer to Question 14

    1. The amount released is the integral of the rate: \(\int_0^550e^{-0.2t}dt=\left[-250e^{-0.2t}\right]_0^5=250(1-e^{-1})\approx\) 158 mg.
    2. The total is an improper integral: \(\int_0^\infty50e^{-0.2t}dt=\displaystyle\lim_{b\to\infty}250(1-e^{-0.2b})=\) 250 mg, because \(e^{-0.2b}\to0\).
    3. The fraction released in 5 hours is \(\dfrac{250(1-e^{-1})}{250}=1-e^{-1}\approx\) 63.2% of the total.
    4. Solve \(250(1-e^{-0.2t})=0.9\times250\): \(1-e^{-0.2t}=0.9\), so \(e^{-0.2t}=0.1\) and \(t=5\ln10\approx\) 11.5 hours.

15. Answer to Question 15

    1. Trapezium: \(\dfrac h2\left[f_0+2(f_1+\dots+f_5)+f_6\right]=1\times\left[0+2(0.6+1.1+1.5+1.2+0.7)+0\right]=\) 10.2 m\(^2\).
    2. Simpson: \(\dfrac23\left[0+4(0.6)+2(1.1)+4(1.5)+2(1.2)+4(0.7)+0\right]=\dfrac23[2.4+2.2+6+2.4+2.8]=\dfrac23(15.8)\approx\) 10.53 m\(^2\).
    3. Flow \(=\) area \(\times\) speed: trapezium \(10.2\times0.8=\) 8.16 m\(^3\)/s; Simpson \(10.53\times0.8\approx\) 8.43 m\(^3\)/s.
    4. Simpson's rule fits parabolas through triples of points and is usually more accurate for smooth shapes like a river bed, while the trapezium rule underestimates where the bed is curved outwards (concave down).

16. Answer to Question 16

    1. On the left, \(0.1P\left(1-\dfrac P{100}\right)=\dfrac{P(100-P)}{1000}\), so the reciprocal is \(\dfrac{1000}{P(100-P)}\). On the right, \(\dfrac1P+\dfrac1{100-P}=\dfrac{(100-P)+P}{P(100-P)}=\dfrac{100}{P(100-P)}\); multiplying by 10 gives the same expression.
    2. Integrating term by term, \(\int10\left(\dfrac1P+\dfrac1{100-P}\right)dP=10\left(\ln P-\ln(100-P)\right)=10\ln\dfrac P{100-P}\) for \(0\lt P\lt 100\). Evaluating between the limits gives the result.
    3. \(T=10\left[\ln\dfrac{50}{50}-\ln\dfrac{20}{80}\right]=10\left[0-\ln\dfrac14\right]=10\ln4\approx\) 13.9 years.
    4. As \(P_2\to100\), \(\dfrac{P_2}{100-P_2}\to\infty\), so its logarithm tends to infinity. The integral diverges: the population approaches the carrying capacity 100 but never reaches it in finite time.

17. Answer to Question 17

    1. Discs of radius \(\dfrac1x\) give \(\pi y^2dx=\dfrac{\pi}{x^2}dx\). Then \(V=\pi\left[-\dfrac1x\right]_1^\infty=\pi(0+1)=\) \(\pi\).
    2. Since \(\sqrt{1+x^{-4}}\ge1\), the integrand is at least \(\dfrac1x\).
    3. \(\int_1^b\dfrac{dx}x=\ln b\to\infty\) as \(b\to\infty\). So the lower bound diverges, and \(S=\infty\).
    4. A finite volume can be filled with a finite amount of paint (\(\pi\) cubic units), but the horn has an infinite surface. The paint inside covers the infinite inside surface in a layer thinner and thinner as you go along. A real paint layer cannot become arbitrarily thin, so the paradox is a feature of the idealised mathematics.

18. Answer to Question 18

    1. \(I_0=\left[-e^{-t}\right]_0^\infty=0+1=\) 1.
    2. Take \(u=t^n\), \(v'=e^{-t}\): \(I_n=\left[-t^ne^{-t}\right]_0^\infty+n\int_0^\infty t^{n-1}e^{-t}dt=0+nI_{n-1}\).
    3. \(I_1=1\cdot1=\) 1, \(I_2=2\cdot1=\) 2, \(I_3=3\cdot2=\) 6, \(I_4=4\cdot6=\) 24. These are factorials, so \(I_n=n!\).
    4. The boundary term \(\left[-t^ne^{-t}\right]_0^\infty\) vanishes at \(\infty\) because exponentials grow faster than any power (l'Hôpital's rule applied \(n\) times), and at 0 because \(t^n=0\) for \(n\ge1\).

19. Answer to Question 19

    1. A disc at height \(y\) has radius \(x=\sqrt y\), so area \(\pi y\). Then \(V=\pi\int_0^hy\,dy=\dfrac{\pi h^2}2\).
    2. \(V(9)=\dfrac{81\pi}2\approx\) 127 ml.
    3. \(127.2\div5\approx\) 25.4 seconds.
    4. Half the capacity is \(\dfrac{81\pi}4=\dfrac\pi2h^2\), so \(h^2=\dfrac{81}2\) and \(h=\dfrac9{\sqrt2}\approx\) 6.36 cm. From \(\dfrac{dV}{dt}=\pi h\dfrac{dh}{dt}=5\) we get \(\dfrac{dh}{dt}=\dfrac5{\pi h}\approx\) 0.250 cm/s.

20. Answer to Question 20

    1. Since \(f\) is continuous, the fundamental theorem says that \(\dfrac d{dx}\int_0^xf(t)\,dt=f(x)\). Differentiating both sides of the equation, \(f'(x)=0+f(x)=f(x)\). Setting \(x=0\), the integral is zero, so \(f(0)=1\).
    2. Separating, \(\dfrac{f'}f=1\), so \(\ln f=x+c\) and \(f=Ae^x\). The condition \(f(0)=1\) gives \(A=1\), so \(f(x)=e^x\).
    3. Substituting back: \(1+\int_0^xe^tdt=1+(e^x-1)=e^x\), which equals \(f(x)\). The answer checks.
    4. \(f_1=1+\int_0^x1\,dt=1+x\), then \(f_2=1+x+\dfrac{x^2}2\) and \(f_3=1+x+\dfrac{x^2}2+\dfrac{x^3}6\). Each step adds the next term of the Maclaurin series of \(e^x\), so \(f_n\to e^x\), in agreement with part (b).