Integrating powers and standard functions Problem set B One question at a time

Find \(\int x^{\frac{3}{4}}\,dx\).

• Add 1 to the power: \(\frac{3}{4}+1=\frac{7}{4}\).
• Divide by the new power: \(1\div\frac{7}{4}=\frac{4}{7}\).
• \(\int x^{\frac{3}{4}}\,dx=\frac{4}{7}x^{\frac{7}{4}} + c\)

Find \(\int x^{-\frac{5}{2}}\,dx\).

• Add 1 to the power: \(-\frac{5}{2}+1=-\frac{3}{2}\).
• Divide by the new power: \(1\div(-\frac{3}{2})=-\frac{2}{3}\).
• \(\int x^{-\frac{5}{2}}\,dx=-\frac{2}{3}x^{-\frac{3}{2}} + c\)

Find \(\int 10x^{\frac{3}{5}}\,dx\).

• Add 1 to the power: \(\frac{3}{5}+1=\frac{8}{5}\).
• Divide by the new power: \(10\div\frac{8}{5}=\frac{25}{4}\).
• \(\int 10x^{\frac{3}{5}}\,dx=\frac{25}{4}x^{\frac{8}{5}} + c\)

Find \(\int \frac{5}{8}\sqrt{x}\,dx\).

• Write the root as a power: \(\sqrt{x}=x^{\frac{1}{2}}\).
• Add 1 to the power: \(\frac{1}{2}+1=\frac{3}{2}\).
• Divide by the new power: \(\frac{5}{8}\div\frac{3}{2}=\frac{5}{12}\).
• \(\int \frac{5}{8}\sqrt{x}\,dx=\frac{5}{12}x^{\frac{3}{2}} + c\)

Find \(\int \frac{12}{\sqrt[4]{x}}\,dx\).

• Write as a negative power: \(\frac{12}{\sqrt[4]{x}}=12x^{-\frac{1}{4}}\).
• Add 1 to the power: \(-\frac{1}{4}+1=\frac{3}{4}\).
• Divide by the new power: \(12\div\frac{3}{4}=16\).
• \(\int \frac{12}{\sqrt[4]{x}}\,dx=16x^{\frac{3}{4}} + c\)

Find y in terms of x.

1. (a) \(\frac{dy}{dx}=9\cos x\)

   • The integral of \(\cos x\) is \(\sin x\).
   • \(y=9\sin x+c\)

2. (b) \(\frac{dy}{dx}=4\sin x\)

   • The integral of \(\sin x\) is \(-\cos x\).
   • \(y=-4\cos x+c\)

A curve has \(\frac{dy}{dx}=5e^x\) and passes through \((0,\,2)\). Find its equation.

• \(y=5e^x+c\)
• \(2=5e^0+c=5+c, \quad c=-3\)
• \(y=5e^x-3\)

A curve has \(\frac{dy}{dx}=\frac{5}{x}\) for \(x>0\) and passes through \((1,\,-2)\). Find its equation.

• \(y=5\ln|x|+c\)
• \(-2=5\ln 1+c=c\), since \(\ln 1=0\)
• \(y=5\ln x-2\)

Find y, given that \(\frac{dy}{dx}=2\cos x+6e^x-3\).

• \(\int 2\cos x\,dx=2\sin x\): the integral of \(\cos x\) is \(\sin x\).
• \(\int 6e^x\,dx=6e^x\), \(\int -3\,dx=-3x\).
• \(y=2\sin x+6e^x-3x+c\). Add one c for the whole sum.

The depth L metres of water in a harbour basin changes at the rate \(\frac{dL}{dt}=\frac{2}{\sqrt[3]{t}}\), where \(t\) is the time in hours and \(t>0\). When \(t=1\), \(L=5\) m.

1. (a) Find an expression for \(L\) in terms of \(t\).

   • Write as a power: \(\frac{2}{\sqrt[3]{t}}=2t^{-\frac{1}{3}}\).
   • \(L=\int 2t^{-\frac{1}{3}}\,dt=3t^{\frac{2}{3}}+c\)
   • Put in the point: \(5=3(1)^{\frac{2}{3}}+c\), so \(c=2\).

2. (b) Find \(L\) when \(t=27\), to 3 significant figures.

   • \(L=3(27)^{\frac{2}{3}}+2=29\) m

Oil leaks from a tank. The rate of leak is \(\frac{dV}{dt}=15e^{-3t}\) litres per minute, where \(V\) litres is the amount leaked and \(t\) is the time in minutes. When \(t=0\), \(V=0\).

1. (a) Find \(V\) in terms of \(t\).

   • \(V=\int 15e^{-3t}\,dt=-5e^{-3t}+c\): divide by \(-3\), the number in front of t.
   • Put in the point: \(0=-5e^0+c\), so \(c=5\).

2. (b) Find \(V\) when \(t=1\), to 3 significant figures.

   • \(V=-5e^{-3\times 1}+5=4.75\) litres

A curve has \(\frac{dy}{dx}=12x^{\frac{1}{3}}-\frac{6}{x}+\frac{3}{\sqrt[3]{x}}\) for \(x>0\), and passes through \((8,\,30)\). Find the equation of the curve, with the constant to 3 significant figures.

• Write \(\frac{3}{\sqrt[3]{x}}\) as \(3x^{-\frac{1}{3}}\), then integrate each term.
• \(y=9x^{\frac{4}{3}}-6\ln x+\frac{9}{2}x^{\frac{2}{3}}+c\)
• Put in the point: \(30=9(8)^{\frac{4}{3}}-6\ln 8+\frac{9}{2}(8)^{\frac{2}{3}}+c\).
• \(c=-120\), so \(y=9x^{\frac{4}{3}}-6\ln x+\frac{9}{2}x^{\frac{2}{3}}-120\)

Find \(\int \frac{(\sqrt{x}+2)^2}{x}\,dx\).

• Expand the top: \((\sqrt{x}+2)^2=x+4\sqrt{x}+4\).
• Divide every term by x: \(1+4x^{-\frac12}+\frac{4}{x}\).
• Integrate each term: \(x+8x^{\frac12}+4\ln|x|+c\).