Area between a curve and the y-axis Problem set B Worked solutions

Find the area between the curve \(y=\sqrt{x}\), the y-axis and the lines \(y=3\) and \(y=5\).

• Make x the subject: \(x=y^{2}\)
• Area = \(\int_{3}^{5} y^{2}\,dy\)
• From the calculator: \(\int_{3}^{5} y^{2}\,dy = 32.7\)

Find the area between the curve \(y=x^{3}\), the y-axis and the lines \(y=1\) and \(y=64\).

• Make x the subject: \(x=\sqrt[3]{y}\)
• Area = \(\int_{1}^{64} \sqrt[3]{y}\,dy\)
• From the calculator: \(\int_{1}^{64} \sqrt[3]{y}\,dy = 191\)

Find the area between the curve \(y=e^{0.5x}\), the y-axis and the lines \(y=2\) and \(y=7\).

• \(y=e^{0.5x}\Rightarrow \ln y=0.5x\Rightarrow x=2\ln y\)
• Area = \(\int_{2}^{7}2\ln y\,dy\)
• From the calculator: \(\int_{2}^{7}2\ln y\,dy = 14.5\)

Find the area between the curve \(y=\frac{7}{x-3}\), the y-axis and the lines \(y=2\) and \(y=6\).

• Make x the subject: \(x-3=\frac{7}{y}\), so \(x=\frac{7}{y}+3\)
• Area = \(\int_{2}^{6}\left(\frac{7}{y}+3\right)dy\)
• From the calculator: \(\int_{2}^{6}\left(\frac{7}{y}+3\right)dy = 19.7\)

Find the area between the curve \(y=\tan x\), the y-axis and the lines \(y=1.5\) and \(y=5\). Use radians.

• Make x the subject: \(x=\tan^{-1}y\)
• Area = \(\int_{1.5}^{5} \tan^{-1}y\,dy\)
• Radian mode.
• From the calculator: \(\int_{1.5}^{5} \tan^{-1}y\,dy = 4.35\)

Find the area between the curve \(y=5\sin x\), the y-axis and the lines \(y=1\) and \(y=4\), for \(0\le x\le\frac{\pi}{2}\).

• Make x the subject: \(\sin x=\frac{y}{5}\), so \(x=\sin^{-1}\!\left(\frac{y}{5}\right)\)
• Area = \(\int_{1}^{4} \sin^{-1}\!\left(\frac{y}{5}\right)dy\)
• Radian mode.
• From the calculator: \(\int_{1}^{4} \sin^{-1}\!\left(\frac{y}{5}\right)dy = 1.61\)

The side of a grass embankment beside a cycle path is curved, with lengths in metres. The region is bounded by the curve \(y=e^{0.5x}\), the y-axis and the lines \(y=1.5\) and \(y=8\). Find its area in m².

• \(y=e^{0.5x}\Rightarrow \ln y=0.5x\Rightarrow x=2\ln y\)
• Area = \(\int_{1.5}^{8}2\ln y\,dy\)
• From the calculator: \(\int_{1.5}^{8}2\ln y\,dy = 19.1\)

The glass of a lamp shade has a curved cross-section, with lengths in centimetres. Find the area between the curve \(y=\frac{9}{x-1}\), the y-axis and the lines \(y=2\) and \(y=6\), in cm².

• Make x the subject: \(x-1=\frac{9}{y}\), so \(x=\frac{9}{y}+1\)
• Area = \(\int_{2}^{6}\left(\frac{9}{y}+1\right)dy\)
• From the calculator: \(\int_{2}^{6}\left(\frac{9}{y}+1\right)dy = 13.9\)

The diagram shows the curve \(y=\ln x\). The shaded area is \(\int_{1}^{5}\ln x\,dx\).

1. (a) Find the area between the curve \(y=\ln x\), the y-axis and the lines \(y=0\) and \(y=\ln 5\).

   • \(x=e^{y}\)
   • \(\int_{0}^{\ln 5} e^{y}\,dy=\left[e^{y}\right]_{0}^{\ln 5}=5-1=4\)

2. (b) Use the rectangle with corners \((0,0)\) and \((5,\ln 5)\) to show that \(\int_{1}^{5}\ln x\,dx=5\ln 5-4\).

   • Rectangle = shaded area + area to the y-axis: \(5\ln 5=\int_{1}^{5}\ln x\,dx+4\)
   • \(\int_{1}^{5}\ln x\,dx=5\ln 5-4\)

A banner is cut from a region of cloth. The area between the curve \(y=\sqrt{x}\), the y-axis and the lines \(y=2\) and \(y=k\), where \(k>2\), is 25. Find k.

• \(x=y^{2}\)
• Area = \(\int_{2}^{k}y^{2}\,dy=\frac{k^{3}-8}{3}\)
• \(\frac{k^{3}-8}{3}=25\Rightarrow k^{3}=83\)
• k = 4.36