Exercise Set 12 Answers

  1. Determine the following indefinite integrals

    1. \(\int 2x^7\; dx\)
    2. \(\int 5x^2+2x-1 \; dx\)
    3. \(\int \sin(\theta)+\cos(\theta) \; d\theta\)
    4. \(\int 3e^t + \frac{2}{t} \; d t\)
    5. \(\int x^\frac{1}{2} \; dx\)

    Answers:

    1. \(\int 2x^7 \; dx = \frac{1}{4}x^8 + c\)
    2. \(\int 5x^2 + 2x-1\; dx = \frac{5}{3}x^3 + x^2 - x + c\)
    3. \(\int \sin(\theta)+\cos(\theta) \; d\theta = -\cos(\theta) + \sin(\theta) + c\)
    4. \(\int 3e^t + \frac{2}{t} \; dt = 3e^t + 2\ln(t) + c\)
    5. \(\int x^\frac{1}{2} \; dx = \frac{2}{3}x^{\frac{3}{2}} + c\)

  2. Determine the following definite integrals

    1. \(\int_0^2 x^2-2x\; dx\)
    2. \(\int_4^{16} \frac{2}{\sqrt{x}} \; dx\)
    3. \(\int_{-r}^r \frac{mx^2}{2r}\: dx\)
    4. \(\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos(x)\; dx\)
    5. \(\int_{\ln(2)}^{\ln(3)} e^x\; dx\)

    Answers:

    1. \(\int_0^2 x^2-2x\; dx = \left[\frac{x^3}{3} - x^2\right]_0^2 = \left(\frac{2^3}{3} - 2^2\right) - 0 = -\frac{4}{3}\)
    2. \(\int_4^{16} \frac{2}{\sqrt{x}} \; dx = \int_4^{16} 2x^{-\frac{1}{2}}\;dx = \left[4x^{\frac{1}{2}}\right]_4^16 = 16 - 8 = 8\)
    3. Note that the limit, \(r\), appears inside the integral, but it is just a constant. It is explicitly taken out of the integral here \(\int_{-r}^r \frac{mx^2}{2r}\:dx = \frac{m}{2r}\int_{-r}^r x^2 \: dx = \frac{m}{2r}\left[\frac{1}{3}x^3\right]_{-r}^r = \frac{m}{2r}\left(\frac{r^3}{3} + \frac{r^3}{3}\right) = \frac{mr^2}{3}\)
    4. \(\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos(x)\; dx = \left[\sin(x)\right]_{-\frac{\pi}{2}}^\frac{\pi}{2} = \sin(\pi/2) - \sin(\pi/2) = 2\)
    5. \(\int_{\ln(2)}^{\ln(3)} e^x\; dx = \left[e^x\right]_{\ln(2)}^{\ln(3)} = e^{\ln(3)} - e^{\ln(2)} = 3-2 = 1\)

  3. Let \(F(x)\) be an anti-derivative of \(f(x)\). Find an anti-derivative of \(f(ax+b)\) where \(a\) and \(b\) are constants. Hint: use the chain rule on \(F(ax+b)\).

    Answer:

    Let \(u(x)=ax+b\), then the derivative of \(F(ax+b)=F(u)\) is (using the chain rule) \[\frac{dF(ax+b)}{dx}=\frac{dF}{du}\times\frac{du}{dx}=f(u)\times a=af(ax+b)\] where we have used the fact that \(\frac{dF(u)}{du}=f(u)\), since \(F(x)\) is an anti-derivative of \(f(x)\). This means that \[\frac{d\frac{1}{a}F(ax+b)}{dx}=f(ax+b)\] that is, \(\frac{1}{a}F(ax+b)\) is an anti-derivative of \(f(ax+b)\). Hence, \[\int f(ax+b)\; dx = \frac{1}{a}F(ax+b).\]

  4. Use your result to the previous question to find the following indefinite integrals

    1. \(\int (2x+6)^2\; dx\)
    2. \(\int (2x+6)^{99} \; dx\)
    3. \(\int \sin(3x+2) \; dx\)
    4. \(\int \cos(-4x-1) \; dx\)
    5. \(\int \sec^2(9x+2) \; dx\)
    6. \(\int \frac{1}{2x+7}\; dx\)
    7. \(\int e^{-3x+1}\; dx\)
    8. \(\int \frac{1}{(5x+1)^2}\; dx\)
    9. \(\int \sqrt{3x+4}\; dx\)

    Answer:

    1. \(\int (2x+6)^2\; dx=\frac{1}{2}\times\frac{1}{3}(2x+6)^3+C=\frac{1}{6}(2x+6)^3+C\)
    2. \(\int (2x+6)^{99} \; dx=\frac{1}{2}\times\frac{1}{100}(2x+6)^{100}+C=\frac{1}{200}(2x+6)^{100}+C\)
    3. \(\int \sin(3x+2) \; dx=-\frac{1}{3}\cos(3x+2)+C\)
    4. \(\int \cos(-4x-1) \; dx=\frac{1}{4}\sin(-4x-1)+C\)
    5. \(\int \sec^2(9x+2) \; dx=\frac{1}{9}\tan(9x+2)+C\)
    6. \(\int \frac{1}{2x+7}\; dx= \frac{1}{2}\ln(2x+7)+C\)
    7. \(\int e^{-3x+1}\; dx= -\frac{1}{3}e^{-3x+1}+C\)
    8. \(\int \frac{1}{(5x+1)^2}\; dx= \int (5x+1)^{-2}\; dx=\frac{1}{5}\times\frac{1}{-1}(5x+1)^{-1}+C=-\frac{1}{5(5x+1)}+C\)
    9. \(\int \sqrt{3x+4}\; dx= \int (3x+4)^\frac{1}{2}\; dx=\frac{1}{3}\frac{2}{3}(3x+4)^\frac{3}{2}+C=\frac{2}{9}(3x+4)^\frac{3}{2}+C\)

  5. Find the area between the curve and the \(x\)-axis of the function \(f(x)=x^3-x\) and the lines \(x=0\) and \(x=2\).

    Answer:

    We are looking for the area enclosed by the function and the \(x\)-axis, but we cannot simply take the definite integral from \(0\) to \(2\) as the function is negative at some points on this interval.

    The roots of \(f(x)=x^3-x=x(x+1)(x-1)\) are \(-1,0,1\) and these are where the function changes sign. We can plug in values between the roots to see where \(f(x)\) is positive and where it is negative. Since we are interested in the interval from \(0\) to \(2\) we check \(f(0.5)=-3.75\) and \(f(2)=6\). So the area will be given by \[A=\int_0^1 -f(x)\, dx + \int_1^2 f(x)\, dx.\] Hence \[A=\left[-\frac{1}{4}x^4 + \frac{1}{2}x^2\right]_0^1 + \left[\frac{1}{4}x^4 - \frac{1}{2}x^2\right]_1^2 = (\frac{1}{4}-0) + (2 - (-1/4)) = \frac{5}{2}.\]

  6. The velocity \(v\) (\(\text{ms}^{-1}\)) of a particle moving in a straight line at time \(t\) (\(\text{s}\)) is given by: \[v=6t^2-30t-36\quad t>0\] By integrating, find the distance travelled by the particle between \(t=3\) and \(t=5\) seconds.

    Answer:

    \[\int_3^5 6t^2 - 30t - 36\: dt = \left[2t^3 - 15t^2 - 36t\right]_3^5 = -116\] Since we are looking for a distance, we take the absolute value: \(116\text{m}\).

  7. The Root Mean Square (RMS) value of a signal over the interval \(a\) to \(b\) is defined as \[\left(\frac{1}{b-a}\int_{a}^{b} (f(x))^2 \; dx\right)^\frac{1}{2}\] Find the RMS of the voltage signal \[V(t)=3\sin(2t)+5\cos(4t)\] between \(a=0\) and \(b=2\pi\).

    Answer:

    Firstly, multiply out the square term inside the integral, so \[\begin{align*} v(t)^2 &= (3\sin(2t) + 5\cos(4t))(3\sin(2t) + 5\cos(4t))\\ &= 9\sin^2(2t) + 30\sin(2t)\cos(4t) + 25\cos^2(4t) \end{align*}\] So the individual terms in the integral, with constants taken outside of the integral, are \[ \int_0^{2\pi} (v(t))^2\:dt = 9\int_0^{2\pi} \sin^2(2t)\:dt + 30\int_0^{2\pi} \sin(2t)\cos(4t)\:dt + 25 \int_0^{2\pi}\cos^2(4t)\:dt \] We evaluate each of the terms separately: \[ 9\int_0^{2\pi} \sin^2(2t)\:dt = 9\int_0^{2\pi}\frac{1-\cos(4t)}{2}\:dt\] having used the cosine double-angle formula: \(\cos(2\theta) = 1- 2\sin^2(\theta)\). \[ \frac{9}{2}\left[t - \frac{1}{4}\sin(4t)\right]_0^{2\pi} = \frac{9}{2}(2\pi - 0) = 9\pi\] For the second term, we again need a trigonometric identity, in this case the product to sum formula \(\sin(\theta)\cos(\phi) = \frac{1}{2}(\sin(\theta + \phi) + \sin(\theta - \phi))\): \[ 30\int_0^{2\pi} \sin(2t)\cos(4t)\: dt = \frac{30}{2}\int_0^{2\pi} = \sin(6t) + \sin(-2t)\:dt = 15\int_0^{2\pi}\sin(6t) - \sin(2t)\:dt \\= 15\left[-\frac{1}{6}\cos(6t) + \frac{1}{2}\cos(2t)\right]_0^{2\pi} = 0 \] And finally, using \(\cos^2(4t) = \frac{1}{2}(\cos(8t)+1)\): \[ 25\int_0^{2\pi}\cos^2(4t)\: dt = \frac{25}{2}\int_0^{2\pi}\cos(8t) + 1\:dt \\= \frac{25}{2}\left[\frac{1}{8}\sin(8t) + t\right]_0^{2\pi} = 25\pi \] Putting these three terms together: \[ \int_0^{2\pi}(v(t))^2\:dt = 34\pi\\ \left(\frac{1}{2\pi - 0}\int_0^{2\pi}(v(t))^2\:dt\right)^{\frac{1}{2}} = \left(\frac{34\pi}{2\pi}\right)^\frac{1}{2} = \sqrt{17} \]

  8. The volume of revolution of a curve is defined as \[\int_a^b \pi (f(x))^2 \; dx\] Find the volume of revolution of the curve \[f(x)=e^{-2x}\] between \(a=1\) and \(b=2\).

    Answer:

    If \(f(x) = e^{-2x}\) then \((f(x))^2 = e^{-4x}\). So the voume of revolution of the curve \(f(x)\) between \(a=1\) and \(b=2\) is \[ \int_1^2 \pi e^{-4x}\;dx = \pi\left[-\frac{1}{4}e^{-4x}\right]_1^2 = \pi\left(\frac{-e^{-8}}{4} - \frac{-e^{-4}}{4}\right) =0.0141 \text{ to 3d.p.} \]

  9. The points \((1,1)\) and \((2,8)\) on the curve \(y=x^3\) are joined by a straight line. Calculate the area between the line and the curve.

    Answer:

    The area between two curves described by funtions \(f_1(x)\) and \(f_2(x)\) is the area between the function \((f_1(x) - f_2(x))\) and the \(x\)-axis. First, find the equation of the line between the points \((1,1)\) and \((2,8)\): \[ f_1(x) = 7x - 6, \quad \text{since the gradient is } \frac{8-1}{2-1}. \] So the area between the line \(y=f_1(x)\) and the line \(y=f_2(x) = x^3\) is \[\begin{align*} \int_1^2 (f_1(x)-f_2(x))\;dx &= \int_1^2 7x-6-x^3\;dx\\ &= \left[\frac{7}{2}x^2 - 6x - \frac{1}{4}x^4\right]_1^2\\ &= (14-12-4)-(\frac{7}{2} - 6 -\frac{1}{4})\\ &= \frac{3}{4} \end{align*}\]