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

Find the area between the curve \(y=\sqrt{x}\), the y-axis and the lines \(y=2\) and \(y=4\).

• Make x the subject: \(x=y^{2}\)
• Area = \(\int_{2}^{4} y^{2}\,dy\)
• From the calculator: \(\int_{2}^{4} y^{2}\,dy = 18.7\)

Find the area between the curve \(y=x^{3}\), the y-axis and the lines \(y=8\) and \(y=27\).

• Make x the subject: \(x=\sqrt[3]{y}\)
• Area = \(\int_{8}^{27} \sqrt[3]{y}\,dy\)
• From the calculator: \(\int_{8}^{27} \sqrt[3]{y}\,dy = 48.8\)

Find the area between the curve \(y=e^{5x}\), the y-axis and the lines \(y=3\) and \(y=8\).

• \(y=e^{5x}\Rightarrow \ln y=5x\Rightarrow x=\frac{1}{5}\ln y\)
• Area = \(\int_{3}^{8}0.2\ln y\,dy\)
• From the calculator: \(\int_{3}^{8}0.2\ln y\,dy = 1.67\)

Find the area between the curve \(y=\frac{3}{x-2}\), the y-axis and the lines \(y=1\) and \(y=4\).

• Make x the subject: \(x-2=\frac{3}{y}\), so \(x=\frac{3}{y}+2\)
• Area = \(\int_{1}^{4}\left(\frac{3}{y}+2\right)dy\)
• From the calculator: \(\int_{1}^{4}\left(\frac{3}{y}+2\right)dy = 10.2\)

Find the area between the curve \(y=\tan x\), the y-axis and the lines \(y=0.5\) and \(y=3\). Use radians.

• Make x the subject: \(x=\tan^{-1}y\)
• Area = \(\int_{0.5}^{3} \tan^{-1}y\,dy\)
• Radian mode.
• From the calculator: \(\int_{0.5}^{3} \tan^{-1}y\,dy = 2.48\)

Find the area between the curve \(y=2\sin x\), the y-axis and the lines \(y=0.5\) and \(y=1.5\), for \(0\le x\le\frac{\pi}{2}\).

• Make x the subject: \(\sin x=\frac{y}{2}\), so \(x=\sin^{-1}\!\left(\frac{y}{2}\right)\)
• Area = \(\int_{0.5}^{1.5} \sin^{-1}\!\left(\frac{y}{2}\right)dy\)
• Radian mode.
• From the calculator: \(\int_{0.5}^{1.5} \sin^{-1}\!\left(\frac{y}{2}\right)dy = 0.532\)

A skate ramp has a curved side panel, with lengths in metres. The region is bounded by the curve \(y=e^{0.4x}\), the y-axis and the lines \(y=2\) and \(y=9\). Find its area in m².

• \(y=e^{0.4x}\Rightarrow \ln y=0.4x\Rightarrow x=2.5\ln y\)
• Area = \(\int_{2}^{9}2.5\ln y\,dy\)
• From the calculator: \(\int_{2}^{9}2.5\ln y\,dy = 28.5\)

The wall of a plastic funnel is shaped like a curve, with lengths in centimetres. Find the area between the curve \(y=\frac{6}{x-2}\), the y-axis and the lines \(y=3\) and \(y=8\), in cm².

• Make x the subject: \(x-2=\frac{6}{y}\), so \(x=\frac{6}{y}+2\)
• Area = \(\int_{3}^{8}\left(\frac{6}{y}+2\right)dy\)
• From the calculator: \(\int_{3}^{8}\left(\frac{6}{y}+2\right)dy = 15.9\)

The diagram shows the curve \(y=\ln x\). The shaded area is \(\int_{1}^{7}\ln x\,dx\).

1. (a) Find the area between the curve \(y=\ln x\), the y-axis and the lines \(y=0\) and \(y=\ln 7\).

   • \(x=e^{y}\)
   • \(\int_{0}^{\ln 7} e^{y}\,dy=\left[e^{y}\right]_{0}^{\ln 7}=7-1=6\)

2. (b) Use the rectangle with corners \((0,0)\) and \((7,\ln 7)\) to show that \(\int_{1}^{7}\ln x\,dx=7\ln 7-6\).

   • Rectangle = shaded area + area to the y-axis: \(7\ln 7=\int_{1}^{7}\ln x\,dx+6\)
   • \(\int_{1}^{7}\ln x\,dx=7\ln 7-6\)

A sail is cut from a region of cloth. The area between the curve \(y=\sqrt{x}\), the y-axis and the lines \(y=1\) and \(y=k\), where \(k>1\), is 18. Find k.

• \(x=y^{2}\)
• Area = \(\int_{1}^{k}y^{2}\,dy=\frac{k^{3}-1}{3}\)
• \(\frac{k^{3}-1}{3}=18\Rightarrow k^{3}=55\)
• k = 3.80