Definite integrals and area under a curve Problem set B Worked solutions

Evaluate \(\int_{-3}^{2} 3x^4\,dx\).

• Write the integral: \(\int_{-3}^{2} 3x^4\,dx\)
• From the calculator: \(\int_{-3}^{2} 3x^4\,dx\) = 165

Evaluate \(\int_{1}^{4} (7-2t^3)\,dt\).

• Write the integral: \(\int_{1}^{4} (7-2t^3)\,dt\)
• From the calculator: \(\int_{1}^{4} (7-2t^3)\,dt\) = −107

Evaluate \(\int_{1}^{4} 6x^{-3}\,dx\).

• Write the integral: \(\int_{1}^{4} 6x^{-3}\,dx\)
• From the calculator: \(\int_{1}^{4} 6x^{-3}\,dx\) = 2.81

Evaluate \(\int_{0}^{2} (3x+2)^3\,dx\).

• Write the integral: \(\int_{0}^{2} (3x+2)^3\,dx\)
• From the calculator: \(\int_{0}^{2} (3x+2)^3\,dx\) = 340

The diagram shows the curve y = 15 − x². Find the area of the shaded region between x = 2 and x = 3.

• Area = \(\int_{2}^{3} (15-x^2)\,dx\)
• From the calculator: \(\int_{2}^{3} (15-x^2)\,dx\) = 8.67

The diagram shows the curve y = 2 + 8/x². Find the area of the shaded region between x = 2 and x = 5.

• Area = \(\int_{2}^{5} (2+8x^{-2})\,dx\)
• From the calculator: \(\int_{2}^{5} (2+8x^{-2})\,dx\) = 8.4

The diagram shows the curve y = 8/x. Find the area of the shaded region between x = 2 and x = 7.

• Area = \(\int_{2}^{7} \frac{8}{x}\,dx\)
• From the calculator: \(\int_{2}^{7} \frac{8}{x}\,dx\) = 10.0

The diagram shows the curve \(y=\frac{x^2}{3}\) inside the rectangle with corners \((0,0)\), \((6,0)\), \((6,12)\) and \((0,12)\). Find the area of the shaded region above the curve.

• Rectangle: \(6\times12=72\)
• Under the curve: \(\int_0^{6}\frac{x^2}{3}\,dx\) = 24
• Shaded area = \(72-24\) = 48

A pond lies between the x-axis and the curve \(y=6-0.5x^2\), from \(x=0\) to \(x=3\). Lengths are in metres. Find the area of the pond.

• Area = \(\int_{0}^{3} (6-0.5x^2)\,dx\)
• From the calculator: \(\int_{0}^{3} (6-0.5x^2)\,dx\) = 13.5 m²

A news story is shared online at a rate of \(120e^{-0.15t}\) messages per day, where t is the time in days after it is posted. Find the total number of messages in the first 10 days.

• Total \(=\int_0^{10} 120e^{-0.15t}\,dt\)
• From the calculator: \(\int_0^{10} 120e^{-0.15t}\,dt\) = 621, so about 621 messages

The cross-section of a greenhouse is the region between the x-axis (the ground) and the curve \(y=3-\frac{x^2}{3}\), where x and y are in metres. The greenhouse is 12 m long.

1. (a) Find the width of the greenhouse at ground level.

   • \(3-\frac{x^2}{3}=0\), so \(x^2=9\) and \(x=\pm3\)
   • Width \(=3-(-3)=6\) m

2. (b) Find the area of the cross-section.

   • Area \(=\int_{-3}^{3}\left(3-\frac{x^2}{3}\right)dx\)
   • From the calculator: 12 m²

3. (c) Find the volume of air inside the greenhouse.

   • Volume = area × length = \(12\times12\)
   • = 144 m³

The diagram shows the curve \(y=8-\dfrac{3}{x^2}\). It meets the x-axis at the point \((a,0)\).

1. (a) Find the exact value of a.

   • \(8-\frac{3}{x^2}=0\), so \(x^2=\frac{3}{8}\)
   • Take the positive root: \(a=\sqrt{\frac{3}{8}}=0.612\)

2. (b) Find the area under the curve from x = a to x = 2.

   • Area = \(\int_{a}^{2} (8-3x^{-2})\,dx\)
   • From the calculator: \(\int_{0.612}^{2} (8-3x^{-2})\,dx\) = 7.70