Integrals under transformations Problem set B Worked solutions

Given that \(\int_{1}^{6} f(x)\,dx=2.8\), find:

1. (a) \(\int_{1}^{6} 6f(x)\,dx\)

   • Constant multiple: \(6\times 2.8=16.8\)

2. (b) \(\int_{1}^{6} -f(x)\,dx\)

   • Factor −1: \(-(2.8)=−2.8\)

Given that \(\int_{1}^{6} f(x)\,dx=14.5\):

1. (a) find \(\int_{5}^{10} g(x)\,dx\), where \(g(x)=f(x-4)\);

   • The graph moves 4 to the right, so the interval 1 to 6 becomes 5 to 10.
   • The area is unchanged: 14.5

2. (b) find \(\int_{-2}^{3} p(x)\,dx\), where \(p(x)=f(x+3)\).

   • The graph moves 3 to the left, so the interval 1 to 6 becomes -2 to 3.
   • The area is unchanged: 14.5

Given that \(\int_{1}^{5} f(x)\,dx=8.5\), find:

1. (a) \(\int_{1}^{5} [f(x)+4]\,dx\)

   • Width \(5-1=4\)
   • \(8.5+4\times 4=24.5\)

2. (b) \(\int_{1}^{5} [f(x)-3]\,dx\)

   • \(8.5-3\times 4=−3.5\)

Given that \(\int_{0}^{5} f(x)\,dx=6\):

1. (a) the graph of g is the graph of f stretched parallel to the x-axis with scale factor 4, so \(g(x)=f\left(\frac{x}{4}\right)\). Find \(\int_{0}^{20} g(x)\,dx\);

   • Horizontal stretch, scale factor 4: the interval 0 to 5 becomes 0 to 20.
   • \(4\times 6=24\)

2. (b) \(h(x)=f(5x)\). Find \(\int_{0}^{1} h(x)\,dx\).

   • \(f(5x)\) is a stretch with scale factor \(\tfrac15\): the interval becomes 0 to 1.
   • \(6\div 5=1.2\)

Given that \(\int_{0}^{9} f(x)\,dx=12\), the graph of g is obtained from the graph of f by a horizontal stretch with scale factor \(\frac{1}{3}\) followed by a vertical stretch with scale factor \(5\). Find \(\int_{0}^{3} g(x)\,dx\).

• Horizontal factor \(\frac{1}{3}\), vertical factor \(5\).
• \(\frac{1}{3}\times 5\times 12=20\)

Customers enter a shop at the rate of f(t) per hour, where t is the time in hours after the shop opens. In the first 6 hours 210 customers come in, so \(\int_0^6 f(t)\,dt=210\). Find the number of customers in the first 6 hours for each of these:

1. (a) a bigger shop with customer rate 2.5f(t);

   • A constant multiple: 2.5 × 210 = 525

2. (b) a shop with an advert that brings an extra 7 customers every hour: rate f(t) + 7;

   • The extra 7 is added for the whole 6 hours: \(210+7\times 6=252\)

3. (c) a branch that opens 3 hours later with the same pattern, rate f(t − 3), between t = 3 and t = 9.

   • The start is 3 hours later, so the graph moves 3 to the right: the interval 0 to 6 becomes 3 to 9.
   • The area is unchanged: 210

The rate of rainfall during a storm is R(t) millimetres per hour, where t is the time in hours after it starts. Given that \(\int_0^4 R(t)\,dt=1.6\), a model for a longer, heavier storm is \(g(t)=2.5\,R\left(\frac{t}{3}\right)\). Find \(\int_0^{12} g(t)\,dt\), the total rainfall in millimetres over the 12 hours.

• Horizontal stretch \(3\): 4 hours becomes 12 hours.
• Vertical stretch \(2.5\).
• \(3\times 2.5\times 1.6=12\) mm

Given that \(\int_{0}^{5} f(x)\,dx=4.2\) and \(g(x)=3f\left(\frac{x}{2}\right)-1\), find \(\int_{0}^{10} g(x)\,dx\).

• Split: \(3\int_0^{10} f\left(\tfrac{x}{2}\right)dx-\int_0^{10}1\,dx\)
• Stretched part: 2 × 4.2 = 8.4, times 3 gives 25.2
• Constant part: \(-1\times 10=−10\)
• Total: 15.2

The graph of g is obtained from the graph of f by a horizontal stretch with scale factor 3 followed by a vertical stretch with scale factor 4. Given that \(\int_{0}^{12} g(x)\,dx=150\), find \(\int_{0}^{4} f(x)\,dx.\)

• Total factor \(3\times 4=12\)
• \(150=12\times\int_0^4 f(x)\,dx\)
• \(150\div 12=12.5\)

Given that \(\int_{3}^{9} f(x)\,dx=12\), the graph of g is obtained from the graph of f by a horizontal stretch with scale factor \(\frac{1}{3}\), then a vertical stretch with scale factor \(2\), then a translation of 5 units to the left. The interval from 3 to 9 on the graph of f becomes the interval from p to q on the graph of g.

1. (a) Find p and q.

   • Stretch: \(3\to 1\) and \(9\to 3\)
   • Translate 5 to the left: \(p=−4,\ q=−2\)

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

   • The translation does not change the area. \(\frac{1}{3}\times 2\times 12=8\)

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

1. (a) \(\int_{a}^{b} [4f(x)+5]\,dx\)

   • \(\int_a^b 4f(x)\,dx=4X\)
   • \(\int_a^b (5)\,dx=5(b-a)\)
   • Add: \(4X+5(b-a)\)

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

   • The interval becomes a - 3 to b - 3: the shift does not change the integral of f, so that part is X.
   • The constant gives \(-2(b-a)\): \(X-2(b-a)\)