Mixed practice: integrals and areas Problem set A Worked solutions

Find the exact value of \(\int_0^{\ln 5} 4e^{-2x}\,dx\).

• \(\int 4e^{-2x}\,dx=-2e^{-2x}\)
• \(\left[-2e^{-2x}\right]_0^{\ln 5}=-2e^{-2\ln 5}+2\)
• \(e^{-2\ln 5}=5^{-2}=\frac1{25}\)
• \(2\left(1-\frac1{25}\right)=\frac{48}{25}\)

Find the exact value of \(\int_{-12}^{-3}\frac{8}{x}\,dx\), giving your answer as a multiple of a single logarithm.

• \(\int\frac{8}{x}\,dx=8\ln|x|\) (the limits are negative, so use the modulus)
• \(8\ln3-8\ln12=8\ln\frac{3}{12}\)
• \(8\ln\frac14=-8\ln4=-16\ln2\)

Find the exact value of each integral.

1. (a) \(\int_1^6\frac6{3x+2}\,dx\);

   • \(\int\frac6{3x+2}\,dx=2\ln|3x+2|\)
   • \(2\ln20-2\ln5=2\ln4=4\ln2\)

2. (b) \(\int_1^4\frac1{x-7}\,dx\).

   • \(\int\frac1{x-7}\,dx=\ln|x-7|\)
   • \(\ln|-3|-\ln|-6|=\ln3-\ln6=\ln\frac12=-\ln2\)
   • The answer is negative: the curve is below the axis on this interval.

Show that the value of \(\int_k^{3k}\frac1x\,dx\), where \(k>0\), is independent of k.

• \(\int\frac1x\,dx=\ln|x|\), and \(k>0\) so \(\ln|x|=\ln x\)
• \(\ln(3k)-\ln k\)
• \(\ln\frac{3k}{k}=\ln 3\)
• The answer is \(\ln 3\) whatever k is.

In each part, a is positive. Find a.

1. (a) \(\int_0^a(x^2+2x)\,dx=18\);

   • \(\left[\frac{x^3}3+x^2\right]_0^a=\frac{a^3}3+a^2=18\)
   • \(a^3+3a^2-54=0\)
   • Polynomial solver: a = 3 (the other two roots are complex), so a = 3

2. (b) \(\int_a^{2a}(3x+2)\,dx=46.5\).

   • \(\left[\frac32x^2+2x\right]_a^{2a}=(6a^2+4a)-\left(\frac32a^2+2a\right)=\frac92a^2+2a\)
   • \(9a^2+4a-93=0\)
   • Polynomial solver: a = 3 or a = −3.44. Reject the negative: a = 3

Use the properties of integrals.

1. (a) Given that \(\int_1^5f(x)\,dx=9\), find \(\int_1^5(4f(x)-3)\,dx\);

   • \(\int_1^54f\,dx-\int_1^53\,dx=4\times9-3\times(5-1)\)
   • \(36-12=24\)

2. (b) Given that \(\int_0^8f(x)\,dx=18\), find \(\int_0^43f(2x)\,dx\).

   • \(f(2x)\) has half the width of \(f(x)\): \(\int_0^4f(2x)\,dx=\frac12\int_0^8f(x)\,dx=9\)
   • \(3\times9=27\)

Use reverse differentiation.

1. (a) Differentiate \(x\ln(2x)\) with respect to x;

   • Product rule: \(1\times\ln(2x)+x\times\frac1x=\ln(2x)+1\)

2. (b) hence find the exact value of \(\int_1^3\ln(2x)\,dx\).

   • From (a): \(\int(\ln(2x)+1)\,dx=x\ln(2x)\), so \(\int\ln(2x)\,dx=x\ln(2x)-x\)
   • \(\left[x\ln(2x)-x\right]_1^3=(3\ln6-3)-(\ln2-1)\)
   • \(=3\ln6-\ln2-2=2\ln2+3\ln3-2\)

Find the exact value of \(\int_0^2x\sqrt{x^2+5}\,dx\).

• Differentiate \((x^2+5)^{3/2}\): you get \(3x\sqrt{x^2+5}\), so \(\int x\sqrt{x^2+5}\,dx=\frac13(x^2+5)^{3/2}\)
• \(\frac13\cdot9^{3/2}-\frac13\cdot5^{3/2}=\frac{27}3-\frac{5\sqrt5}3\)
• \(=9-\frac{5\sqrt5}{3}\)

The curves \(y=e^{2x}\) and \(y=e^{-x}\) meet at \((0,1)\). Find the exact area of the shaded region, which lies below both curves, above the x-axis, from \(x=-1\) to \(x=1\).

• The lower curve is \(e^{2x}\) for \(x<0\) and \(e^{-x}\) for \(x>0\). Split at x = 0.
• \(\int_{-1}^0e^{2x}\,dx=\left[\frac12e^{2x}\right]_{-1}^0=\frac12-\frac12e^{-2}\)
• \(\int_0^1e^{-x}\,dx=\left[-e^{-x}\right]_0^1=1-e^{-1}\)
• Total: \(\frac32-e^{-1}-\frac12e^{-2}\)

A flower bed is the region enclosed by the curves \(y=e^{-x^2}\) and \(y=e^{-2x}\), where x and y are in metres. The curves meet at A on the y-axis and at B.

1. (a) Find the coordinates of A and B.

   • Solve \(e^{-x^2}=e^{-2x}\) with the calculator: x = 0 or x = 2
   • At x = 0, y = 1. At x = 2, \(y=e^{-4}=0.0183\)

2. (b) Find the area of the flower bed.

   • On \(0\le x\le2\), \(e^{-x^2}\) is above \(e^{-2x}\)
   • From the calculator: \(\int_0^2\left(e^{-x^2}-e^{-2x}\right)dx=0.391\)
   • Area = 0.391 m²

The diagram shows the graphs of \(y=\sin x\) and \(y=\cos x\) for \(0\le x\le\frac\pi2\).

1. (a) Write down the exact coordinates of the point where the graphs meet.

   • \(\sin x=\cos x\) when \(\tan x=1\), so \(x=\frac\pi4\)
   • \(y=\sin\frac\pi4=\frac{\sqrt2}2\)

2. (b) Find the exact area of the shaded region, between the two graphs and the lines \(x=0\) and \(x=\frac\pi2\).

   • Cosine is above sine up to \(\frac\pi4\); sine is above after it. Split there.
   • \(\int_0^{\pi/4}(\cos x-\sin x)\,dx=\left[\sin x+\cos x\right]_0^{\pi/4}=\sqrt2-1\)
   • \(\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx=\left[-\cos x-\sin x\right]_{\pi/4}^{\pi/2}=-1+\sqrt2\)
   • Total: \(2\sqrt2-2\)

Let \(f(x)=\sin(x^2)\) and \(g(x)=\cos(x^2)\), for \(0\le x\le1.5\).

1. (a) Sketch the graphs of f and g on the same axes.

   • Radians mode. Graph both with a window 0 ≤ x ≤ 1.5.
   • From the calculator: the graphs meet at x = 0.886, where y = 0.707

2. (b) Find the area enclosed by the two graphs and the y-axis.

   • g is above f from x = 0 to the crossing
   • From the calculator: \(\int_0^{0.886}\left(\cos(x^2)-\sin(x^2)\right)dx=0.611\)
   • Area = 0.611

Show that the area enclosed by the graph of \(y=\sin3x\) and the x-axis for \(0\le x\le\frac{2\pi}3\) is \(\frac43\).

• \(\sin3x=0\) at \(x=\frac\pi3\): the graph is above the axis before it and below it after.
• \(\int_0^{\pi/3}\sin3x\,dx=\left[-\frac13\cos3x\right]_0^{\pi/3}=\frac13+\frac13=\frac23\)
• \(\int_{\pi/3}^{2\pi/3}\sin3x\,dx=-\frac23\). The area is the positive value \(\frac23\).
• Area \(=\frac23+\frac23=\frac43\). (The plain integral over the whole interval is 0, which is not the area.)

The diagram shows part of the graph of \(y=f(x)\), which crosses the x-axis at a, b, c and d. The diagram is not drawn to scale. Given that \(\int_a^b f(x)\,dx=7\), \(\int_a^d f(x)\,dx=5\) and \(\int_a^d|f(x)|\,dx=21\), find

1. (a) \(\int_b^d f(x)\,dx\);

   • \(\int_a^d f=\int_a^b f+\int_b^d f\), so \(\int_b^d f=5-7=−2\)

2. (b) \(\int_a^c f(x)\,dx\);

   • Let the areas be 7 (a to b, above), q (b to c, below) and r (c to d, above).
   • \(p+q+r=21\) and \(p-q+r=5\). Subtract: \(2q=16\), so q = 8
   • \(\int_a^c f=7-8=−1\)

3. (c) the area of the region between the curve and the x-axis from b to c.

   • The area from b to c is q = 8