Application of Integration

Finding a Function from a Known Rate of Change

• Concept: If we know the rate of change of a function (its derivative) and a boundary condition (a specific value of the function at a particular point), we can find the original function using integration.

• Method:

  1. Integrate the rate of change function (the derivative) to find the general form of the original function.
  2. Use the boundary condition to find the constant of integration (\(C\)).
  3. Write the specific equation for the function.

• Example: Given \(\frac{dy}{dx} = 2x\) and \(y(1) = 5\), find \(y(x)\).

  1. Integrate: \(y = \int 2x \, dx = x^2 + C\).
  2. Apply boundary condition: \(5 = (1)^2 + C \Rightarrow C = 4\).
  3. Therefore, \(y = x^2 + 4\).

Calculation of the Area of a Region Under a Curve

• Concept: The definite integral of a function \(f(x)\) from \(a\) to \(b\) represents the area between the curve \(y = f(x)\), the x-axis, and the vertical lines \(x = a\) and \(x = b\).
• Formula: Area \(= \int_a^b f(x) \, dx\)
• Note: Areas below the x-axis are considered negative. To find the total area, consider the absolute value of the integral in those regions.
• Example: Find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 2\).
  Area \(= \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}\)

Areas Between Curves

• Concept: To find the area between two curves, \(y = f(x)\) and \(y = g(x)\), from \(x = a\) to \(x = b\), integrate the difference between the two functions.
• Formula: Area \(= \int_a^b |f(x) - g(x)| \, dx\). It’s important to take the absolute value to ensure area is positive. Alternatively, split the integral into sections where you integrate (top function - bottom function).

• Steps:

  1. Find the points of intersection of the two curves by solving \(f(x) = g(x)\). These points will be your limits of integration (\(a\) and \(b\)).
  2. Determine which function is greater (i.e., lies above) the other in the interval \([a, b]\).
  3. Integrate the difference: \(\int_a^b (f(x) - g(x)) \, dx\), where \(f(x)\) is the upper function and \(g(x)\) is the lower function.

• Example: Find the area between \(y = x^2\) and \(y = x\) from \(x = 0\) to \(x = 1\).

  • Here, \(x > x^2\) on \([0,1]\)
    Area \(= \int_0^1 (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{1}{2} - \frac{1}{3} \right) - (0 - 0) = \frac{1}{6}\)

Average Value of a Function

• Concept: The average value of a function \(f(x)\) over the interval \([a, b]\) is the height of a rectangle with base \((b-a)\) that has the same area as the area under the curve \(y = f(x)\) from \(a\) to \(b\).
• Formula: Average Value \(= \frac{1}{b - a} \int_a^b f(x) \, dx\)
• Example: Find the average value of \(f(x) = x^2\) on the interval \([0, 3]\).
  Average Value \(= \frac{1}{3 - 0} \int_0^3 x^2 \, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^3 = \frac{1}{3} \left( \frac{27}{3} - 0 \right) = \frac{1}{3} (9) = 3\)

Rates of Change and Accumulation

• Concept: Integration can be used to find the total change in a quantity given its rate of change.
• Application: If \(v(t)\) represents the velocity of an object at time \(t\), then \(\int_a^b v(t) \, dt\) represents the displacement (change in position) of the object from time \(t = a\) to \(t = b\).
• Total Distance: To find the total distance traveled, integrate the absolute value of the velocity function: \(\int_a^b |v(t)| \, dt\).

Other Situations

• Integration can be applied to a variety of problems involving accumulation, such as:

  • Population growth: Given a rate of population increase, find the total population at a future time.
  • Fluid flow: Given the rate of fluid flow into a tank, find the total volume of fluid in the tank at a given time.

  • Marginal cost/revenue: Given the marginal cost or revenue function, find the total cost or revenue by integrating.

  • Example: The rate of water leaking from a tank is given by \(L'(t) = 5e^{-0.2t}\) liters per minute. How much water leaks out in the first 10 minutes?

    • Amount leaked \(= \int_0^{10} 5e^{-0.2t} dt = 5 \int_0^{10} e^{-0.2t} dt = 5\left[ \frac{e^{-0.2t}}{-0.2} \right]_0^{10} = -25[e^{-2} - e^{0}] = -25[e^{-2} - 1] \approx 21.62\) liters.