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

Evaluate \(\int_{-2}^{1} 5x^4\,dx\).

• Write the integral: \(\int_{-2}^{1} 5x^4\,dx\)
• From the calculator: \(\int_{-2}^{1} 5x^4\,dx\) = 33

Evaluate \(\int_{1}^{3} (4-5t^3)\,dt\).

• Write the integral: \(\int_{1}^{3} (4-5t^3)\,dt\)
• From the calculator: \(\int_{1}^{3} (4-5t^3)\,dt\) = −92

Evaluate \(\int_{2}^{5} 8x^{-3}\,dx\).

• Write the integral: \(\int_{2}^{5} 8x^{-3}\,dx\)
• From the calculator: \(\int_{2}^{5} 8x^{-3}\,dx\) = 0.84

Evaluate \(\int_{1}^{3} (2x-1)^3\,dx\).

• Write the integral: \(\int_{1}^{3} (2x-1)^3\,dx\)
• From the calculator: \(\int_{1}^{3} (2x-1)^3\,dx\) = 78

The diagram shows the curve y = 12 − x². Find the area of the shaded region between x = 1 and x = 3.

• Area = \(\int_{1}^{3} (12-x^2)\,dx\)
• From the calculator: \(\int_{1}^{3} (12-x^2)\,dx\) = 15.3

The diagram shows the curve y = 3 + 6/x². Find the area of the shaded region between x = 1 and x = 3.

• Area = \(\int_{1}^{3} (3+6x^{-2})\,dx\)
• From the calculator: \(\int_{1}^{3} (3+6x^{-2})\,dx\) = 10

The diagram shows the curve y = 6/x. Find the area of the shaded region between x = 1 and x = 5.

• Area = \(\int_{1}^{5} \frac{6}{x}\,dx\)
• From the calculator: \(\int_{1}^{5} \frac{6}{x}\,dx\) = 9.66

The diagram shows the curve \(y=\frac{x^2}{4}\) inside the rectangle with corners \((0,0)\), \((6,0)\), \((6,9)\) and \((0,9)\). Find the area of the shaded region above the curve.

• Rectangle: \(6\times9=54\)
• Under the curve: \(\int_0^{6}\frac{x^2}{4}\,dx\) = 18
• Shaded area = \(54-18\) = 36

A flower bed lies between the x-axis and the curve \(y=5-0.25x^2\), from \(x=0\) to \(x=4\). Lengths are in metres. Find the area of the flower bed.

• Area = \(\int_{0}^{4} (5-0.25x^2)\,dx\)
• From the calculator: \(\int_{0}^{4} (5-0.25x^2)\,dx\) = 14.7 m²

A medicine is absorbed into the blood at a rate of \(40e^{-0.25t}\) milligrams per hour, where t is the time in hours after the dose. Find the total amount absorbed in the first 6 hours.

• Total \(=\int_0^{6} 40e^{-0.25t}\,dt\)
• From the calculator: \(\int_0^{6} 40e^{-0.25t}\,dt\) = 124 mg

The cross-section of a tunnel is the region between the x-axis (the ground) and the curve \(y=8-0.5x^2\), where x and y are in metres. The tunnel is 150 m long.

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

   • \(8-0.5x^2=0\), so \(x^2=16\) and \(x=\pm4\)
   • Width \(=4-(-4)=8\) m

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

   • Area \(=\int_{-4}^{4}(8-0.5x^2)\,dx\)
   • From the calculator: 42.7 m²

3. (c) Find the volume of rock removed to make the tunnel.

   • Volume = area × length = \(42.67\times150\)
   • = 6400 m³

The diagram shows the curve \(y=5-\dfrac{2}{x^2}\). It meets the x-axis at the point \((a,0)\).

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

   • \(5-\frac{2}{x^2}=0\), so \(x^2=\frac{2}{5}\)
   • Take the positive root: \(a=\sqrt{\frac{2}{5}}=0.632\)

2. (b) Find the area under the curve from x = a to x = 3.

   • Area = \(\int_{a}^{3} (5-2x^{-2})\,dx\)
   • From the calculator: \(\int_{0.632}^{3} (5-2x^{-2})\,dx\) = 9.34