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

The diagram shows the graph of \(y = x^{2} - 4 x - 5\). The shaded region is bounded by the curve, the x-axis and the lines \(x = 0\) and \(x = 4\). Find the area of the shaded region.

• \(\int x^{2} - 4 x - 5\,dx = \frac{x^{3}}{3} - 2 x^{2} - 5 x\)
• \(\int_{0}^{4}\left(x^{2} - 4 x - 5\right)dx = - \frac{92}{3}\): below the axis, area \(\frac{92}{3}\)
• Total area = \(\frac{92}{3}\) ≈ 30.7

The diagram shows the graph of \(y = x^{3} - 4 x^{2} + 3 x\). The curve crosses the x-axis at \(x = 0, 1, 3\). Find the total area of the shaded region between \(x = 0\) and \(x = 3\).

• The curve crosses the x-axis at x = 1 between the limits, so split there.
• \(\int x^{3} - 4 x^{2} + 3 x\,dx = \frac{x^{4}}{4} - \frac{4 x^{3}}{3} + \frac{3 x^{2}}{2}\)
• \(\int_{0}^{1}\left(x^{3} - 4 x^{2} + 3 x\right)dx = \frac{5}{12}\): above the axis, area \(\frac{5}{12}\)
• \(\int_{1}^{3}\left(x^{3} - 4 x^{2} + 3 x\right)dx = - \frac{8}{3}\): below the axis, area \(\frac{8}{3}\)
• Total area = \(\frac{37}{12}\) ≈ 3.08

The diagram shows the graph of \(y = x^{4} - 2 x^{3} - 5 x^{2} + 6 x\) between \(x = -2\) and \(x = 3\). Use your calculator to find the total area of the shaded regions.

• Area = \(\int_{-2}^{3}\left|x^{4} - 2 x^{3} - 5 x^{2} + 6 x\right|dx\)
• From the calculator: \(22.9\)
• Area = 22.9

The diagram shows the graph of \(y = \sin x + \cos x\) for \(\frac{\pi}{2} \le x \le 2 \pi\). Use your calculator, in radians, to find the total area between the curve and the x-axis.

• Set the calculator to radians.
• Area = \(\int_{\frac{\pi}{2}}^{2 \pi}\left|\sin x + \cos x\right|dx\)
• From the calculator: \(3.66\)
• Area = 3.66

Evaluate \(\int_{1}^{5}\sqrt{4x+5}\,dx\).

• Let \(u = 4x+5\). Then \(\frac{du}{dx} = 4\), so \(dx = \frac{du}{4}\)
• Limits: \(x = 1\) gives \(u = 9\); \(x = 5\) gives \(u = 25\)
• \(\int_{1}^{5}\sqrt{4x+5}\,dx = \frac{1}{4}\int_{9}^{25}u^{1/2}\,du = \frac{1}{4}\left[\frac{2}{3}u^{3/2}\right]_{9}^{25}\)
• \(= \frac{1}{4}\left(\frac23(25)^{3/2} - \frac23(9)^{3/2}\right) = \frac{1}{4}\times\frac23\left(125 - 27\right)\)
• \(= \frac{49}{3}\)

The curve \(y = 4\sin 5x\) crosses the x-axis at \(x = 0\) and \(x = \frac{\pi}{5}\). Find the exact area enclosed by this part of the curve and the x-axis.

• \(y = 0\) when \(\sin 5x = 0\), so the lobe runs from \(x = 0\) to \(x = \frac{\pi}{5}\)
• Area = \(\int_0^{\pi/5} 4\sin 5x\,dx = \left[-\frac{4}{5}\cos 5x\right]_0^{\pi/5}\)
• \(= -\frac{4}{5}\cos\pi - \left(-\frac{4}{5}\cos 0\right) = \frac{4}{5} + \frac{4}{5}\)
• \(= \frac{8}{5}\)

Find the value of a, where \(a > 9\), such that \(\int_{9}^{a} \frac{6}{\sqrt{x}}\,dx = 18\).

• \(\int_{9}^{a} \frac{6}{\sqrt x}\,dx = \int_{9}^{a} 6x^{-1/2}\,dx = \left[12\sqrt x\right]_{9}^{a} = 12\sqrt a - 36\)
• \(12\sqrt a - 36 = 18\), so \(12\sqrt a = 54\)
• \(\sqrt a = \frac{9}{2}\), so \(a = \frac{81}{4}\)

A curve has equation \(y = - x^{2} + 2 x + 8\).

1. (a) Find the x-coordinates of the points where the curve crosses the x-axis.

   • Solve \(- x^{2} + 2 x + 8 = 0\) with the polynomial solver, or factorise: \(- \left(x - 4\right) \left(x + 2\right) = 0\)
   • \(x = -2\) or \(x = 4\)

2. (b) Find the area enclosed by the curve and the x-axis.

   • \(\int - x^{2} + 2 x + 8\,dx = - \frac{x^{3}}{3} + x^{2} + 8 x\)
   • \(\int_{-2}^{4}\left(- x^{2} + 2 x + 8\right)dx = 36\): above the axis, area \(36\)
   • Total area = \(36\)

A flower bed has a curved edge modelled by \(y = \sqrt{\ln x}\), where x and y are in metres. The bed is bounded by this curve, the x-axis and the lines \(x = 3\) and \(x = 9\). Use your calculator to find the area of the flower bed.

• Area = \(\int_{3}^{9}\sqrt{\ln x}\,dx\). The curve is above the axis for \(x \ge 1\), so no modulus is needed
• No antiderivative we can write: use the calculator
• From the calculator: \(7.90\)
• Area = 7.90 m²

The arch of a footbridge is modelled by the curve \(y = \ln(17 - x^2)\), where x and y are in metres and the x-axis is the ground. Use your calculator to find the area of the opening under the arch.

• The curve meets the x-axis where \(\ln(17 - x^2) = 0\), so \(17 - x^2 = 1\) and \(x = \pm4\)
• Area = \(\int_{-4}^{4}\ln(17 - x^2)\,dx\)
• From the calculator: \(18.5\)
• Area = 18.5 m²

The depth of a canal, \(d\) metres, at a distance \(x\) metres from its centre line is modelled by \(d=1.5e^{-x^2/12}\) for \(x\ge 0\). The cross-sectional area of water from the centre line to the point \(x=k\) is 3.5 m². Find k, to 3 significant figures.

• \(\int_0^{k}1.5e^{-x^2/12}\,dx=3.5\)
• No antiderivative: solve for k with the calculator
• From the calculator: \(k=2.88\)
• 3.5/1.5 = 2.33, and k is larger than that, as it should be

The diagram shows the graph of \(y = x^{3} + 2 x^{2} - 5 x - 6\), which crosses the x-axis at \(x = −3, −1, 2\). The shaded region is bounded by the curve and the x-axis.

1. (a) Evaluate \(\int_{-3}^{-1} y\,dx\) and \(\int_{-1}^{2} y\,dx\).

   • \(\int y\,dx = \frac{x^{4}}{4} + \frac{2 x^{3}}{3} - \frac{5 x^{2}}{2} - 6 x\)
   • \(\int_{-3}^{-1} y\,dx = \frac{16}{3}\)
   • \(\int_{-1}^{2} y\,dx = - \frac{63}{4}\)

2. (b) Hence find the total area of the shaded region.

   • One integral is positive and one is negative: the regions are on opposite sides of the axis
   • Area = \(\left|\frac{16}{3}\right| + \left|- \frac{63}{4}\right|\)
   • Area = \(\frac{253}{12}\) ≈ 21.1