Integrals under transformations Problem set A Worked solutions

Given that \(\int_{0}^{7} f(x)\,dx=6.4\), find:

1. (a) \(\int_{0}^{7} 5f(x)\,dx\)

   • Constant multiple: \(5\times 6.4=32\)

2. (b) \(\int_{0}^{7} -f(x)\,dx\)

   • Factor −1: \(-(6.4)=−6.4\)

Given that \(\int_{3}^{10} f(x)\,dx=26\):

1. (a) find \(\int_{8}^{15} g(x)\,dx\), where \(g(x)=f(x-5)\);

   • The graph moves 5 to the right, so the interval 3 to 10 becomes 8 to 15.
   • The area is unchanged: 26

2. (b) find \(\int_{1}^{8} p(x)\,dx\), where \(p(x)=f(x+2)\).

   • The graph moves 2 to the left, so the interval 3 to 10 becomes 1 to 8.
   • The area is unchanged: 26

Given that \(\int_{2}^{7} f(x)\,dx=11\), find:

1. (a) \(\int_{2}^{7} [f(x)+6]\,dx\)

   • Width \(7-2=5\)
   • \(11+6\times 5=41\)

2. (b) \(\int_{2}^{7} [f(x)-1]\,dx\)

   • \(11-1\times 5=6\)

Given that \(\int_{0}^{4} f(x)\,dx=7\):

1. (a) the graph of g is the graph of f stretched parallel to the x-axis with scale factor 3, so \(g(x)=f\left(\frac{x}{3}\right)\). Find \(\int_{0}^{12} g(x)\,dx\);

   • Horizontal stretch, scale factor 3: the interval 0 to 4 becomes 0 to 12.
   • \(3\times 7=21\)

2. (b) \(h(x)=f(2x)\). Find \(\int_{0}^{2} h(x)\,dx\).

   • \(f(2x)\) is a stretch with scale factor \(\tfrac12\): the interval becomes 0 to 2.
   • \(7\div 2=3.5\)

Given that \(\int_{0}^{6} f(x)\,dx=15\), the graph of g is obtained from the graph of f by a horizontal stretch with scale factor \(2\) followed by a vertical stretch with scale factor \(\frac{1}{3}\). Find \(\int_{0}^{12} g(x)\,dx\).

• Horizontal factor \(2\), vertical factor \(\frac{1}{3}\).
• \(2\times \frac{1}{3}\times 15=10\)

A tank fills at the rate of f(t) litres per minute, where t is the time in minutes after 09:00. In the first 20 minutes the tank takes in 340 litres, so \(\int_0^{20} f(t)\,dt=340\). Find the volume taken in during the first 20 minutes by each of these:

1. (a) a second tank with fill rate 3f(t);

   • A constant multiple: 3 × 340 = 1020

2. (b) a third tank with fill rate f(t) + 4 (a small extra pump runs all the time);

   • The extra 4 is added for the whole 20 minutes: \(340+4\times 20=420\)

3. (c) a fourth tank with the same pattern of filling, but starting 10 minutes later, with fill rate f(t − 10), between t = 10 and t = 30.

   • The start is 10 minutes later, so the graph moves 10 to the right: the interval 0 to 20 becomes 10 to 30.
   • The area is unchanged: 340

The power used by a factory is E(t) kilowatts, where t is the time in hours after the start of a shift. Given that \(\int_0^6 E(t)\,dt=18\), a new model for a slower shift is \(g(t)=1.5\,E\left(\frac{t}{2}\right)\). Find \(\int_0^{12} g(t)\,dt\), the energy used in kilowatt-hours over the 12 hour shift.

• Horizontal stretch \(2\): 6 hours becomes 12 hours.
• Vertical stretch \(1.5\).
• \(2\times 1.5\times 18=54\) kWh

Given that \(\int_{0}^{4} f(x)\,dx=7.5\) and \(g(x)=2f\left(\frac{x}{3}\right)+4\), find \(\int_{0}^{12} g(x)\,dx\).

• Split: \(2\int_0^{12} f\left(\tfrac{x}{3}\right)dx+\int_0^{12}4\,dx\)
• Stretched part: 3 × 7.5 = 22.5, times 2 gives 45
• Constant part: \(4\times 12=48\)
• Total: 93

The graph of g is obtained from the graph of f by a horizontal stretch with scale factor 5 followed by a vertical stretch with scale factor 2. Given that \(\int_{0}^{15} g(x)\,dx=84\), find \(\int_{0}^{3} f(x)\,dx.\)

• Total factor \(5\times 2=10\)
• \(84=10\times\int_0^3 f(x)\,dx\)
• \(84\div 10=8.4\)

Given that \(\int_{1}^{5} f(x)\,dx=10\), the graph of g is obtained from the graph of f by a horizontal stretch with scale factor \(\frac{1}{2}\), then a vertical stretch with scale factor \(3\), then a translation of 4 units to the right. The interval from 1 to 5 on the graph of f becomes the interval from p to q on the graph of g.

1. (a) Find p and q.

   • Stretch: \(1\to 0.5\) and \(5\to 2.5\)
   • Translate 4 to the right: \(p=4.5,\ q=6.5\)

2. (b) Find \(\int_{p}^{q} g(x)\,dx\).

   • The translation does not change the area. \(\frac{1}{2}\times 3\times 10=15\)

Given that \(\int_a^b f(x)\,dx=X\), find, in terms of a, b and X:

1. (a) \(\int_{a}^{b} [3f(x)-2]\,dx\)

   • \(\int_a^b 3f(x)\,dx=3X\)
   • \(\int_a^b (-2)\,dx=-2(b-a)\)
   • Add: \(3X-2(b-a)\)

2. (b) \(\int_{a+4}^{b+4} [f(x-4)+6]\,dx\)

   • The interval becomes a + 4 to b + 4: the shift does not change the integral of f, so that part is X.
   • The constant gives \(6(b-a)\): \(X+6(b-a)\)