Mixed practice: integrals and areas Problem set B Worked solutions

Find the exact value of \(\int_0^{\ln 3} 6e^{-3x}\,dx\).

• \(\int 6e^{-3x}\,dx=-2e^{-3x}\)
• \(\left[-2e^{-3x}\right]_0^{\ln 3}=-2e^{-3\ln 3}+2\)
• \(e^{-3\ln 3}=3^{-3}=\frac1{27}\)
• \(2\left(1-\frac1{27}\right)=\frac{52}{27}\)

Find the exact value of \(\int_{-10}^{-2}\frac{15}{x}\,dx\), giving your answer as a multiple of a single logarithm.

• \(\int\frac{15}{x}\,dx=15\ln|x|\) (the limits are negative, so use the modulus)
• \(15\ln2-15\ln10=15\ln\frac{2}{10}\)
• \(15\ln\frac15=-15\ln5\)

Find the exact value of each integral.

1. (a) \(\int_0^4\frac6{2x+1}\,dx\);

   • \(\int\frac6{2x+1}\,dx=3\ln|2x+1|\)
   • \(3\ln9-3\ln1=3\ln9=6\ln3\)

2. (b) \(\int_0^3\frac5{x-8}\,dx\).

   • \(\int\frac5{x-8}\,dx=5\ln|x-8|\)
   • \(5\ln|-5|-5\ln|-8|=5\ln5-5\ln8=5\ln\frac58\)
   • The answer is negative: the curve is below the axis on this interval.

Show that the value of \(\int_k^{5k}\frac1x\,dx\), where \(k>0\), is independent of k.

• \(\int\frac1x\,dx=\ln|x|\), and \(k>0\) so \(\ln|x|=\ln x\)
• \(\ln(5k)-\ln k\)
• \(\ln\frac{5k}{k}=\ln 5\)
• The answer is \(\ln 5\) whatever k is.

In each part, a is positive. Find a.

1. (a) \(\int_0^a(3x^2+2x)\,dx=80\);

   • \(\left[x^3+x^2\right]_0^a=a^3+a^2=80\)
   • \(a^3+a^2-80=0\)
   • Polynomial solver, coefficients 1, 1, 0, −80: a = 4 (the other two roots are complex), so a = 4

2. (b) \(\int_a^{2a}(2x+3)\,dx=60\).

   • \(\left[x^2+3x\right]_a^{2a}=(4a^2+6a)-(a^2+3a)=3a^2+3a\)
   • \(3a^2+3a-60=0\)
   • Polynomial solver: a = 4 or a = −5. Reject the negative: a = 4

Use the properties of integrals.

1. (a) Given that \(\int_2^6f(x)\,dx=7\), find \(\int_2^6(2f(x)+5)\,dx\);

   • \(\int_2^62f\,dx+\int_2^65\,dx=2\times7+5\times(6-2)\)
   • \(14+20=34\)

2. (b) Given that \(\int_0^6f(x)\,dx=15\), find \(\int_0^34f(2x)\,dx\).

   • \(f(2x)\) has half the width of \(f(x)\): \(\int_0^3f(2x)\,dx=\frac12\int_0^6f(x)\,dx=7.5\)
   • \(4\times7.5=30\)

Use reverse differentiation.

1. (a) Differentiate \(x\ln(3x)\) with respect to x;

   • Product rule: \(1\times\ln(3x)+x\times\frac1x=\ln(3x)+1\)

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

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

Find the exact value of \(\int_0^3x\sqrt{x^2+7}\,dx\).

• Differentiate \((x^2+7)^{3/2}\): you get \(3x\sqrt{x^2+7}\), so \(\int x\sqrt{x^2+7}\,dx=\frac13(x^2+7)^{3/2}\)
• \(\frac13\cdot16^{3/2}-\frac13\cdot7^{3/2}=\frac{64}3-\frac{7\sqrt7}3\)
• \(=\frac{64-7\sqrt7}{3}\)

The curves \(y=e^{3x}\) and \(y=e^{-2x}\) 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 \(x=2\).

• The lower curve is \(e^{3x}\) for \(x<0\) and \(e^{-2x}\) for \(x>0\). Split at x = 0.
• \(\int_{-1}^0e^{3x}\,dx=\left[\frac13e^{3x}\right]_{-1}^0=\frac13-\frac13e^{-3}\)
• \(\int_0^2e^{-2x}\,dx=\left[-\frac12e^{-2x}\right]_0^2=\frac12-\frac12e^{-4}\)
• Total: \(\frac56-\frac13e^{-3}-\frac12e^{-4}\)

The outline of a pond is the region enclosed by the curves \(y=e^{-2x^2}\) and \(y=e^{-3x}\), 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 \(e^{-2x^2}=e^{-3x}\) with the calculator: x = 0 or x = 1.5
   • At x = 0, y = 1. At x = 1.5, \(y=e^{-4.5}=0.0111\)

2. (b) Find the area of the pond.

   • On \(0\le x\le1.5\), \(e^{-2x^2}\) is above \(e^{-3x}\)
   • From the calculator: \(\int_0^{1.5}\left(e^{-2x^2}-e^{-3x}\right)dx=0.295\)
   • Area = 0.295 m²

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

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 \(x=\pi\).

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

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

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

   • Radians mode. Graph both with a window 0 ≤ x ≤ 1.2.
   • From the calculator: the graphs meet at x = 0.681, where y = 0.894

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: \(\int_0^{0.681}\left(\cos(x^2)-2\sin(x^2)\right)dx=0.459\)
   • Area = 0.459

Show that the area enclosed by the graph of \(y=\cos2x\) and the x-axis for \(0\le x\le\frac\pi2\) is 1.

• \(\cos2x=0\) at \(x=\frac\pi4\): the graph is above the axis before it and below it after.
• \(\int_0^{\pi/4}\cos2x\,dx=\left[\frac12\sin2x\right]_0^{\pi/4}=\frac12\)
• \(\int_{\pi/4}^{\pi/2}\cos2x\,dx=-\frac12\). The area is the positive value \(\frac12\).
• Area \(=\frac12+\frac12=1\). (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 \(\int_a^b f(x)\,dx=6\), \(\int_a^d f(x)\,dx=2\) and \(\int_a^d|f(x)|\,dx=20\), find

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

   • \(\int_a^d f=\int_a^b f+\int_b^d f\), so \(\int_b^d f=2-6=−4\)

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

   • Let the areas be 6 (a to b, above), q (b to c, below) and r (c to d, above).
   • \(p+q+r=20\) and \(p-q+r=2\). Subtract: \(2q=18\), so q = 9
   • \(\int_a^c f=6-9=−3\)

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 = 9