Unit: Integration & Accumulation of Change
Chapter: Properties of Integrals & Techniques
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:
- Linearity of Integrals & Constant multiple rules.
- Bounds, Limits & Reversing the Bounds.
- Integrals of Symmetric & Odd Functions.
- Integrals of Periodic Functions & Absolute Values.
Linearity of Integration & Constant Multiple Rules
- Linearity of Integrals:
- The linearity property states that the integral of a sum or difference of functions is equal to the sum or difference of their integrals.
- Mathematically, if f(x) and g(x) are integrable functions, and a and b are constants, then: ∫[af(x) + bg(x)] dx = a∫f(x) dx + b∫g(x) dx.
- In simpler terms, you can integrate each function separately and then combine the results using addition or subtraction.
- Constant Multiple Rule:
- The constant multiple rules state that the integral of a constant time of a function is equal to the constant times of the integral of the function.
- Mathematically, if f(x) is an integrable function and c is a constant, then: ∫cf(x) dx = c∫f(x) dx.
- This rule allows you to bring constants outside of the integral sign.
Bounds, Limits & Reversing the Bounds:
- Bounds of Integration:
- When evaluating definite integrals, the bounds of integration specify the interval over which the integration is performed.
- The lower bound represents the starting point of the interval, and the upper bound represents the endpoint.
- For example, in the definite integral ∫[a, b] f(x) dx, 'a' is the lower bound, and 'b' is the upper bound.
- Limits of Integration:
- Integrals can be evaluated over finite intervals or unbounded intervals.
- Finite intervals have definite limits of integration, where the lower and upper bounds are specific numbers.
- Unbounded intervals can extend to infinity (∞) or negative infinity (-∞), denoted by ±∞.
- For example, ∫[0, ∞] f(x) dx represents integration from 0 to infinity.
- Reversing the Bounds:
- Reversing the bounds in an integral change the sign of the result.
- Mathematically, if a and b are the original bounds of integration, then: ∫[b, a] f(x) dx = -∫[a, b] f(x) dx.
- This property arises from the fact that reversing the bounds changes the orientation of the integration interval, leading to a negative sign in the result.
Integrals of Symmetric & Odd functions: –
- Integrals of Symmetric Functions:
- A function is symmetric about the y-axis if it has the property that for every value of x in its domain, the corresponding y-values on opposite sides of the y-axis are equal.
- When integrating a symmetric function over an interval symmetric about the y-axis, the result is zero.
- Mathematically, if f(x) is a symmetric function and the interval of integration is [-a, a], then: ∫[-a, a] f(x) dx = 0.
- This property arises because the positive and negative areas on either side of the y-axis cancel each other out.
- Integrals of Odd Functions:
- An odd function is symmetric about the origin (0, 0), meaning that for every value of x in its domain, f(x) = -f(-x).
- When integrating an odd function over an interval symmetric about the origin, the result is also zero.
- Mathematically, if f(x) is an odd function and the interval of integration is [-a, a], then: ∫[-a, a] f(x) dx = 0.
- This property arises because the positive and negative areas on either side of the origin cancel each other out.
Integration of Rational Function & Improper Integrals:
- Integration of Rational Functions:
- A rational function is a ratio of two polynomial functions. The process of integrating a rational function involves finding an antiderivative or indefinite integral of the function.
- To integrate a rational function, various techniques are used, including polynomial long division, partial fractions decomposition, and substitution.
- Polynomial long division is used to express the rational function as a sum of a polynomial and a proper fraction, which can be integrated separately.
- Partial fractions decomposition is a method used to decompose a rational function into simpler fractions, making it easier to integrate.
- After decomposition, the resulting fractions can be integrated using basic integration rules.
- Substitution is often employed to simplify the integral by replacing a variable with a new variable or expression.
- Improper Integrals:
- An improper integral is an integral with infinite limits or an integrand that has discontinuities within the interval of integration.
- Improper integrals are evaluated by taking limits as one or both of the bounds approach infinity or a point of discontinuity.
- There are two types of improper integrals: a. Type 1: Infinite Intervals: In these integrals, one or both of the bounds of integration are ±∞. b. Type 2: Discontinuous Integrand: In these integrals, the integrand has a discontinuity within the interval of integration.
- To evaluate improper integrals, techniques such as limit evaluation, breaking the integral into several parts, or using comparison tests are applied.
- The limits involved in improper integrals ensure that the integral converges to a finite value or diverges to infinity.
Integration of Periodic Function & Absolute values
- Integration of Periodic Functions:
- A periodic function is a function that repeats itself after a certain interval called a period.
- When integrating a periodic function over one period, the result is equal to the integral over any other period.
- Mathematically, if f(x) is a periodic function with period P, then for any real numbers a and b: ∫[a, a + P] f(x) dx = ∫[b, b + P] f(x) dx.
- This property arises from the fact that the positive and negative areas of the periodic function cancel each other out over one period.
- It allows us to evaluate the integral of a periodic function by considering a single period, simplifying the calculation.
- Integration of Absolute Values:
- When integrating a function involving absolute values, the integral can be related to the original function by splitting the integral at the points where the function changes sign.
- Mathematically, if f(x) is a function, then: ∫[a, b] |f(x)| dx = ∫[a, c] f(x) dx + ∫[c, b] -f(x) dx, where c is a point in the interval [a, b] where f(x) changes sign.
- This property arises because the absolute value function removes the sign of the original function, and integrating the negative part cancels out the negative areas of the function.
- It allows us to simplify the integration of functions involving absolute values by considering the different intervals where the function is positive or negative separately.
Example: – Using the Linearity of Integrals and Constant Multiple Rule
Find the integral of the function F(x) = 3x2 + 2sin(x) – 4cos(x) over the interval [0, π].
Solution:
To find the integral of F(x) over the given interval, we can apply the linearity of integrals by integrating each term separately and then summing them.
∫[0, π] F(x) dx = ∫[0, π] (3x2) dx + ∫[0, π] (2sin(x)) dx – ∫[0, π] (4cos(x)) dx
Using the power rule of integration, the first term becomes:
= [x3] from 0 to π
= π3 – 0
= π3
The second term involves integrating sin(x), which is a known function:
= [-2cos(x)] from 0 to π
= -2cos(π) + 2cos(0)
= -2(-1) + 2(1)
= 4
Similarly, for the third term involving cos(x), we have:
= [-4sin(x)] from 0 to π
= -4sin(π) + 4sin(0)
= -4(0) + 4(0)
= 0
Summing up the individual integrals, we get:
∫[0, π] F(x) dx = π3 + 4 + 0
= π3 + 4
So, the value of the integral of F(x) over the interval [0, π] is π3 + 4.
Example 2: – Integration of a Rational Function
Find the integral of the function f(x) = (x2 + 3x + 2) / (x + 2) dx.
Solution:
To integrate the rational function f(x), we can apply the technique of partial fractions decomposition.
First, we divide the numerator by the denominator:
X2 + 3x + 2 = (x + 1)(x + 2)
Now, we express the rational function as a sum of simpler fractions:
f(x) = (x2 + 3x + 2) / (x + 2) = (x + 1) + (1 / (x + 2))
Integrating each term separately, we have:
∫ f(x) dx = ∫ (x + 1) dx + ∫ (1 / (x + 2)) dx
Using the power rule of integration, the first term becomes:
= (x2 / 2 + x) + C1
The second term involves the natural logarithm function:
= ln(|x + 2|) + C2
So, the integral of the rational function f(x) is:
∫ f(x) dx = (x2 / 2 + x) + ln(|x + 2|) + C
Where C1 and C2 are constants of integration.
Key Points
- Linearity: The integral of a sum or difference of functions is the sum or difference of their integrals.
- Constant Multiple Rule: A constant factor can be pulled out of the integral.
- Bounds: Integrals have upper and lower bounds that define the interval of integration.
- Reversing Bounds: Reversing the bounds of integration changes the sign of the result.
- Symmetric Functions: Integrating a symmetric function over a symmetric interval yields zero.
- Odd Functions: Integrating an odd function over a symmetric interval also results in zero.
- Periodic Functions: Integrating a periodic function over one period gives the same result for any period.
- Absolute Values: Integrals involving absolute values can be split at points where the function changes sign.
- Power Rule: The integral of x^n is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
- Substitution: Substituting a new variable can simplify the integral.
- Partial Fractions: Decomposing a rational function into simpler fractions allows for easier integration.
- Improper Integrals: Integrals with infinite limits or discontinuities are evaluated using limits.