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.