Accumulated Change Over An Interval

Unit: Integration & Accumulation of Change

Chapter: Accumulation Change over an Interval

Reference: – Riemann sums, Definite integrals, Fundamental theorem, Antiderivatives, Area under a curve, Accumulation functions, Average value of functions, Mean value theorem for Integrals, Properties & Estimation, Trapezoidal rule, Simpson's Rule, Application & Motion problems.

 

After studying this chapter, you should be able to:

  • Introduction to Riemann sums & Definite Integral.
  • Fundamental theorem of Calculus.
  • Properties of Definite Integrals & Integration technique.
  • Examples & Applications

 

Introduction to Riemann Sums

 

Riemann sums are a calculus method used to approximate the area under a curve by dividing the interval into smaller subintervals and summing the areas of corresponding rectangles or trapezoids. They provide an approximation of the definite integral of a function.

Here are the key aspects of Riemann sums:

  1. Partitioning the Interval: To apply Riemann sums, the interval over which the function is integrated is divided into smaller subintervals. This is done by selecting a partition, which is a set of points that divide the interval into subintervals. The partition is typically denoted by P and can be represented as P = {x0, x1, x2, …, xn}, where x0 is the left endpoint of the interval and xn is the right endpoint.
  2. Subintervals: Each subinterval is determined by consecutive points in the partition. The subintervals have a width determined by the difference between two consecutive points. For example, the first subinterval is [x0, x1], the second is [x1, x2], and so on.
  3. Sample Points: Riemann sums require the selection of sample points within each subinterval. The sample points are used to determine the height of the rectangles or trapezoids. Commonly used methods for selecting sample points include the left endpoint, right endpoint, midpoint, or any arbitrary point within the subinterval.
  4. Types of Riemann Sums: The choice of sample points and the shape of the rectangles or trapezoids determine the type of Riemann sum. There are four commonly used types:
  • Left Riemann Sum: The left endpoint of each subinterval is used as the sample point. The area of each rectangle is determined by multiplying the function value at the left endpoint by the width of the subinterval.
  • Right Riemann Sum: The right endpoint of each subinterval is used as the sample point. The area of each rectangle is determined by multiplying the function value at the right endpoint by the width of the subinterval.
  • Midpoint Riemann Sum: The midpoint of each subinterval is used as the sample point. The area of each rectangle is determined by multiplying the function value at the midpoint by the width of the subinterval.
  • Trapezoidal Riemann Sum: The sample points are chosen as the endpoints of each subinterval. The area of each trapezoid is determined by calculating the average of the function values at the endpoints and multiplying it by the width of the subinterval.
  1. Approximating the Definite Integral: The Riemann sum is obtained by summing the areas of the rectangles or trapezoids corresponding to each subinterval. As the number of subintervals increases, the Riemann sum becomes a better approximation of the definite integral of the function over the given interval.
  2. Notation: Riemann sums are commonly denoted using sigma notation (∑). The general form of the Riemann sum for a function f(x) on the interval [a, b] is:

R = ∑[f(xi*)Δxi]

where xi* represents the sample point within each subinterval, and Δxi is the width of the subinterval.

Riemann Sums & Definite Integral:

 

Here's how the definite integral is used to calculate accumulated change over an interval:

  1. Interval: Consider an interval [a, b] on the x-axis. This interval represents the period over which you want to determine the accumulated change.
  2. Function: Suppose you have a function f(x) defined over the interval [a, b]. This function may represent a rate of change, such as velocity, or any other quantity that varies for x.
  3. Definite Integral: The definite integral of the function f(x) over the interval [a, b] is written as:

∫[a to b] f(x) dx

The dx represents an infinitesimally small change in x, indicating that we are summing up the contributions of the function over the entire interval.

  1. Accumulated Change: The value of the definite integral ∫[a to b] f(x) dx gives you the accumulated change of the quantity represented by the function f(x) over the interval [a, b]. It represents the total net change of the quantity during that interval.
  2. Positive and Negative Areas: The definite integral takes into account the positive and negative areas between the curve and the x-axis. If the function f(x) is above the x-axis, the area contributes positively to the accumulated change. Conversely, if the function is below the x-axis, the area contributes negatively.
  3. Interpretation: Depending on the context, the accumulated change calculated using the definite integral can have various interpretations. For example, if the function represents velocity, the definite integral gives you the displacement or the net change in position over the interval. If the function represents a rate of production, the definite integral gives you the total quantity produced over the interval.

 

Fundamental Theorem of Calculus: –

 

  1. First Part of the Fundamental Theorem of Calculus: The first part of the theorem states that if F(x) is an antiderivative (or indefinite integral) of a function f(x) on an interval [a, b], then the definite integral of f(x) from a to b is given by:

∫[a to b] f(x) dx = F(b) – F(a)

In other words, if you can find an antiderivative of the function f(x), evaluating the definite integral of f(x) over an interval [a, b] is equivalent to subtracting the antiderivative values at the endpoints.

  1. Interpretation of the First Part: This interpretation shows that the definite integral of a function f(x) represents the accumulated change of the antiderivative F(x) over the interval [a, b]. It measures the net change in the antiderivative value between the endpoints.
  2. Second Part of the Fundamental Theorem of Calculus: The second part of the theorem states that if F(x) is continuous on an interval [a, b] and differentiable on the open interval (a, b), and if F'(x) = f(x) for all x in (a, b), then:

d/dx ∫[a to x] f(t) dt = f(x)

In simpler terms, this part of the theorem states that if you differentiate the definite integral of a function f(x) for x, you get back the original function f(x).

  1. Interpretation of the Second Part: The second part of the Fundamental Theorem of Calculus provides a powerful connection between integration and differentiation. It allows us to find antiderivatives by evaluating definite integrals. It also means that the derivative of an accumulated change (definite integral) for the upper limit of integration is equal to the rate of change (instantaneous value) of the function being integrated.

 

Properties of Definite Integrals & Integration Technique:

 

  1. Linearity: The definite integral has the property of linearity, meaning that it distributes over addition and scalar multiplication. Specifically, for functions f(x) and g(x) and constants c and d, the following properties hold:
    • ∫[a to b] (cf(x) + dg(x)) dx = c∫[a to b] f(x) dx + d∫[a to b] g(x) dx
  2. Interval Splitting: The interval of integration can be split into smaller intervals, and the definite integral over the whole interval is equal to the sum of the integrals over the individual subintervals. This property is useful when dealing with functions that have discontinuities or different behaviors over different intervals.
  3. Changing Limits of Integration: If the limits of integration are interchanged, the sign of the definite integral changes. That is, if a < b, then:
    • ∫[a to b] f(x) dx = -∫[b to a] f(x) dx
  4. Adding/Subtracting Integrals: The definite integral of the sum or difference of two functions over an interval is equal to the sum or difference of the definite integrals of the individual functions over the same interval. Mathematically, if f(x) and g(x) are functions, then:
    • ∫[a to b] (f(x) + g(x)) dx = ∫[a to b] f(x) dx + ∫[a to b] g(x) dx

Integration Techniques:

  1. Substitution: Substitution is a powerful technique used to simplify integrals by making a substitution to a new variable. This technique involves selecting an appropriate substitution that simplifies the integrand, allowing for easier integration.
  2. Integration by Parts: Integration by parts is a technique based on the product rule for differentiation. It involves selecting parts of the integrand to differentiate and integrate separately, often using the acronym "LIATE" (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to determine which part to differentiate and integrate.
  3. Trigonometric Integrals: Trigonometric integrals involve integrals of trigonometric functions such as sin(x), cos(x), sec(x), etc. These integrals are evaluated using trigonometric identities, trigonometric substitutions, or by rewriting them in terms of exponentials.
  4. Partial Fractions: Partial fractions are a technique used to decompose a rational function into a sum of simpler fractions. This method allows for the integration of the decomposed fractions, which may be easier to handle than the original rational function.
  5. Trigonometric Substitutions: Trigonometric substitutions involve substituting certain trigonometric functions for other variables to simplify integrals. Common substitutions include replacing expressions involving radicals with trigonometric functions.
  6. Other Techniques: Additional integration techniques include completing the square, integration by symmetry, using tables of integrals, and recognizing special functions or forms that have known antiderivatives.

 

Example: – The rate at which water is flowing out of a tank is given by the function R(t) = 3t2 – 6t, where t represents time in seconds and R(t) is measured in liters per second. Find the total amount of water that has flowed out of the tank from t = 1 to t = 5 seconds.

 

Solution:

To find the total amount of water that has flowed out of the tank, we need to calculate the definite integral of the rate function R(t) over the interval [1, 5].

 

 (3t2 – 6t) dt

 

Using the power rule of integration, we can integrate each term of the function separately:

 

= [t3] from 1 to 5 – [3t2] from 1 to 5

 

= (53 – 13) – (3(52) – 3(12))

 

= (125 – 1) – (75 – 3)

 

= 124 – 72

 

= 52 liters

 

Therefore, the total amount of water that has flowed out of the tank from t = 1 to t = 5 seconds is 52 liters.

 

Example 2:   A particle moves along the x-axis with a velocity given by v(t) = 4t – 3, where t represents time in seconds and v(t) is measured in meters per second. Find the displacement of the particle during the time interval from t = 2 to t = 6 seconds.

 

Solution:

To find the displacement of the particle, we need to calculate the accumulated change over the interval [2, 6] using the definite integral of the velocity function v(t).

The displacement is given by the definite integral:

Using the power rule of integration, we integrate each term of the function separately:

 

=   4t dt –   3 dt

 

= (62 – 22) – (3(6) – 3(2))

 

= (36 – 4) – (18 – 6)

 

= 32 – 12

 

= 20 meters

 

Therefore, the displacement of the particle during the time interval from t = 2 to t = 6 seconds is 20 meters.

Key Points

  • Accumulation change over an interval involves calculating the total change or accumulated effect of a quantity over a specific interval.

 

  • The definite integral is used to calculate accumulation change by finding the area under a curve between two points.

 

  • The limits of integration represent the interval over which the accumulation change is calculated.

 

  • The definite integral measures the net change of a quantity and takes into account both positive and negative changes.

 

  • The accumulated change is determined by evaluating the definite integral of a function over the given interval.

 

  • Riemann sums provide a numerical approximation of accumulation change by dividing the interval into smaller subintervals.

 

  • Different integration techniques, such as substitution and integration by parts, are used to evaluate definite integrals and calculate accumulation change.

 

  • The Fundamental Theorem of Calculus connects accumulation change and definite integrals, stating that the derivative of the definite integral yields the original function being integrated.

 

  • Linearity is a property of definite integrals, allowing them to distribute over addition and scalar multiplication.

 

  • Interval splitting allows the interval of integration to be divided into smaller intervals, simplifying the calculation of accumulation change.

 

  • Changing the limits of integration leads to a change in the sign of the definite integral, reflecting the reversal of the interval.

 

  • Accumulation change is often interpreted as the total sum, net change, displacement, or total quantity produced over the given interval, depending on the context of the problem.

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