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

The diagram shows the graph of \(y = x^{2} + 2 x - 15\). The shaded region is bounded by the curve, the x-axis and the lines \(x = -2\) and \(x = 2\). Find the area of the shaded region.

• \(\int x^{2} + 2 x - 15\,dx = \frac{x^{3}}{3} + x^{2} - 15 x\)
• \(\int_{-2}^{2}\left(x^{2} + 2 x - 15\right)dx = - \frac{164}{3}\): below the axis, area \(\frac{164}{3}\)
• Total area = \(\frac{164}{3}\) ≈ 54.7

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

• The curve crosses the x-axis at x = 0 between the limits, so split there.
• \(\int x^{3} - 3 x^{2} - 4 x\,dx = \frac{x^{4}}{4} - x^{3} - 2 x^{2}\)
• \(\int_{-1}^{0}\left(x^{3} - 3 x^{2} - 4 x\right)dx = \frac{3}{4}\): above the axis, area \(\frac{3}{4}\)
• \(\int_{0}^{4}\left(x^{3} - 3 x^{2} - 4 x\right)dx = -32\): below the axis, area \(32\)
• Total area = \(\frac{131}{4}\) ≈ 32.8

The diagram shows the graph of \(y = x^{4} - 7 x^{2} + 6 x\) between \(x = -3\) and \(x = 2\). Use your calculator to find the total area of the shaded regions.

• Area = \(\int_{-3}^{2}\left|x^{4} - 7 x^{2} + 6 x\right|dx\)
• From the calculator: \(43.4\)
• Area = 43.4

The diagram shows the graph of \(y = \sin x - \cos x\) for \(0 \le x \le 2 \pi\). Use your calculator, in radians, to find the total area between the curve and the x-axis.

• Set the calculator to radians.
• Area = \(\int_{0}^{2 \pi}\left|\sin x - \cos x\right|dx\)
• From the calculator: \(5.66\)
• Area = 5.66

Evaluate \(\int_{0}^{4}\sqrt{3x+4}\,dx\).

• Let \(u = 3x+4\). Then \(\frac{du}{dx} = 3\), so \(dx = \frac{du}{3}\)
• Limits: \(x = 0\) gives \(u = 4\); \(x = 4\) gives \(u = 16\)
• \(\int_{0}^{4}\sqrt{3x+4}\,dx = \frac{1}{3}\int_{4}^{16}u^{1/2}\,du = \frac{1}{3}\left[\frac{2}{3}u^{3/2}\right]_{4}^{16}\)
• \(= \frac{1}{3}\left(\frac23(16)^{3/2} - \frac23(4)^{3/2}\right) = \frac{1}{3}\times\frac23\left(64 - 8\right)\)
• \(= \frac{112}{9}\)

The curve \(y = 6\sin 4x\) crosses the x-axis at \(x = 0\) and \(x = \frac{\pi}{4}\). Find the exact area enclosed by this part of the curve and the x-axis.

• \(y = 0\) when \(\sin 4x = 0\), so the lobe runs from \(x = 0\) to \(x = \frac{\pi}{4}\)
• Area = \(\int_0^{\pi/4} 6\sin 4x\,dx = \left[-\frac{6}{4}\cos 4x\right]_0^{\pi/4}\)
• \(= -\frac{6}{4}\cos\pi - \left(-\frac{6}{4}\cos 0\right) = \frac{6}{4} + \frac{6}{4}\)
• \(= 3\)

Find the value of a, where \(a > 4\), such that \(\int_{4}^{a} \frac{10}{\sqrt{x}}\,dx = 30\).

• \(\int_{4}^{a} \frac{10}{\sqrt x}\,dx = \int_{4}^{a} 10x^{-1/2}\,dx = \left[20\sqrt x\right]_{4}^{a} = 20\sqrt a - 40\)
• \(20\sqrt a - 40 = 30\), so \(20\sqrt a = 70\)
• \(\sqrt a = \frac{7}{2}\), so \(a = \frac{49}{4}\)

A curve has equation \(y = - x^{2} - 2 x + 15\).

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

   • Solve \(- x^{2} - 2 x + 15 = 0\) with the polynomial solver, or factorise: \(- \left(x - 3\right) \left(x + 5\right) = 0\)
   • \(x = -5\) or \(x = 3\)

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

   • \(\int - x^{2} - 2 x + 15\,dx = - \frac{x^{3}}{3} - x^{2} + 15 x\)
   • \(\int_{-5}^{3}\left(- x^{2} - 2 x + 15\right)dx = \frac{256}{3}\): above the axis, area \(\frac{256}{3}\)
   • Total area = \(\frac{256}{3}\) ≈ 85.3

A drainage channel has a cross-section bounded by the curve \(y = \sqrt{\ln x}\), the x-axis and the lines \(x = 2\) and \(x = 7\), where x and y are in metres. Use your calculator to find the area of the cross-section.

• Area = \(\int_{2}^{7}\sqrt{\ln x}\,dx\). 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: \(5.97\)
• Area = 5.97 m²

The roof of a tunnel is modelled by the curve \(y = \ln(26 - x^2)\), where x and y are in metres and the x-axis is the road. Use your calculator to find the area of the tunnel's cross-section.

• The curve meets the x-axis where \(\ln(26 - x^2) = 0\), so \(26 - x^2 = 1\) and \(x = \pm5\)
• Area = \(\int_{-5}^{5}\ln(26 - x^2)\,dx\)
• From the calculator: \(27.2\)
• Area = 27.2 m²

The thickness of a snowdrift, \(t\) metres, at a distance \(x\) metres from its crest is modelled by \(t=1.8e^{-x^2/10}\) for \(x\ge 0\). The area of the cross-section from the crest to the point \(x=k\) is 4 m². Find k, to 3 significant figures.

• \(\int_0^{k}1.8e^{-x^2/10}\,dx=4\)
• No antiderivative: solve for k with the calculator
• From the calculator: \(k=2.82\)
• 4/1.8 = 2.22, and k is larger than that, as it should be

The diagram shows the graph of \(y = x^{3} - 4 x^{2} - 7 x + 10\), which crosses the x-axis at \(x = −2, 1, 5\). The shaded region is bounded by the curve and the x-axis.

1. (a) Evaluate \(\int_{-2}^{1} y\,dx\) and \(\int_{1}^{5} y\,dx\).

   • \(\int y\,dx = \frac{x^{4}}{4} - \frac{4 x^{3}}{3} - \frac{7 x^{2}}{2} + 10 x\)
   • \(\int_{-2}^{1} y\,dx = \frac{99}{4}\)
   • \(\int_{1}^{5} y\,dx = - \frac{160}{3}\)

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 = \(\left|\frac{99}{4}\right| + \left|- \frac{160}{3}\right|\)
   • Area = \(\frac{937}{12}\) ≈ 78.1