Definite Integral - An Introduction

Definite Integral - An introduction

The concept of definite integral is an essential part of calculus.
It helps us to find the area under a curve.
A definite integral gives a numerical value for the area by evaluating the antiderivative of a function.
It is denoted by the symbol ∫ and has an interval over which we are finding the area.
The definite integral is used in various fields such as physics, engineering, and economics. Example: Find the area under the curve y = x^2 between x = 0 and x = 2. Solution:
First, we need to find the antiderivative of the function y = x^2.
The antiderivative of x^2 is (1/3)x^3.
Now, we substitute the values x = 0 and x = 2 into the antiderivative:
- A = ∫(0 to 2) x^2 dx = [(1/3)(2)^3] - [(1/3)(0)^3]
- A = (8/3) - 0 = 8/3 square units. Equation: The definite integral is defined as: ∫(a to b) f(x) dx = F(b) - F(a) where F(x) is the antiderivative of f(x).

Properties of Definite Integrals

The definite integral has certain properties that make it a powerful tool in calculus.
Linearity property: ∫(a to b) [f(x) + g(x)] dx = ∫(a to b) f(x) dx + ∫(a to b) g(x) dx
Constant multiple property: ∫(a to b) kf(x) dx = k∫(a to b) f(x) dx, where k is a constant.
Additivity property: ∫(a to b) f(x) dx + ∫(b to c) f(x) dx = ∫(a to c) f(x) dx
Invariance property: ∫(a to a) f(x) dx = 0
Reversal property: ∫(b to a) f(x) dx = -∫(a to b) f(x) dx Example: Evaluate ∫(0 to 2) (2x + 3) dx. Solution:
Apply the linearity property: ∫(0 to 2) (2x + 3) dx = ∫(0 to 2) 2x dx + ∫(0 to 2) 3 dx
Evaluate each integral separately: ∫(0 to 2) 2x dx = [x^2] from 0 to 2 = (2^2) - (0^2) = 4
∫(0 to 2) 3 dx = [3x] from 0 to 2 = (3(2)) - (3(0)) = 6
Add the results: ∫(0 to 2) (2x + 3) dx = 4 + 6 = 10

Definite Integral of a Constant Function

The definite integral of a constant function over an interval [a, b] is equal to the product of the constant and the length of the interval.
Example: ∫(1 to 4) 3 dx = 3(x) | from 1 to 4 = 3(4) - 3(1) = 9

Definite Integral of a Power Function

The definite integral of a power function can be calculated using the power rule.
The power rule states that if f(x) = x^n, where n is any real number except -1, then ∫ f(x) dx = (1/(n+1))x^(n+1) + C
Example: ∫(0 to 3) x^3 dx = (1/4)x^4 | from 0 to 3 = (1/4)(3^4) - (1/4)(0^4) = 9/4

Definite Integral of Exponential Functions

Definite integrals of exponential functions can be evaluated using substitution or by using properties of exponential functions.
Example: ∫(0 to 2) e^x dx
- Using substitution, let u = x, then du = dx, and the integral becomes ∫ e^u du = e^u | from 0 to 2 = e^2 - e^0 = e^2 - 1
- Using properties, the integral of e^x is e^x itself, so ∫(0 to 2) e^x dx = [e^x] from 0 to 2 = e^2 - e^0 = e^2 - 1

Definite Integral of Trigonometric Functions

The definite integral of trigonometric functions can be evaluated using trigonometric identities and properties.
Example: ∫(0 to π/2) sin(x) dx = [-cos(x)] from 0 to π/2 = -cos(π/2) - (-cos(0)) = -1 - (-1) = 0

Definite Integral of Odd and Even Functions

An odd function satisfies f(-x) = -f(x), while an even function satisfies f(-x) = f(x).
The definite integral of an odd function from -a to a is always 0.
The definite integral of an even function from -a to a is twice the value of the integral from 0 to a.
Example: ∫(-2 to 2) x^3 dx = 0 (since x^3 is an odd function)

Definite Integral as the Net Change

The definite integral can be interpreted as the net change of a function between two points.
Positive values of the definite integral represent net accumulation or increase.
Negative values of the definite integral represent net depletion or decrease.
Example: If a function represents the rate of population growth, the definite integral of the function over a certain time interval would give the net change in the population during that interval.

Applications of Definite Integrals

Definite integrals have numerous applications in various fields.
They can be used to calculate areas and volumes of irregular shapes.
They are used in physics to find displacement, velocity, and acceleration.
They can be used in economics to measure total revenue and cost.
They are used in statistics to calculate probabilities and expected values.

Definite Integral of Composite Functions

The definite integral of a composite function f(g(x)) over an interval [a, b] can be evaluated using the substitution method.
Example: ∫(0 to 1) e^(x^2) 2x dx
- Let u = x^2, then du = 2x dx
- Substitute into the integral: ∫(0 to 1) e^u du = [e^u] from 0 to 1 = e^1 - e^0 = e - 1

Definite Integral of Piecewise Functions

Piecewise functions are functions defined by different expressions in different intervals.
To find the definite integral of a piecewise function, evaluate the integral separately in each interval.
Example: ∫(-2 to 2) |x| dx
- In the interval [-2, 0], |x| = -x, so ∫(-2 to 0) -x dx = [-x^2/2] from -2 to 0 = 0 - (-2^2/2) = 2
- In the interval [0, 2], |x| = x, so ∫(0 to 2) x dx = [x^2/2] from 0 to 2 = (2^2/2) - (0^2/2) = 2
- Add the results: ∫(-2 to 2) |x| dx = 2 + 2 = 4

Riemann Sum and Definite Integral

Riemann Sum is a method to approximate the area under a curve using rectangles.
It is represented by ∑(n) f(x_k) Δx.
Definite integral is the limit of the Riemann Sum as the width of the rectangles approaches zero.
It is denoted by ∫(a to b) f(x) dx. Example:
Let’s find the definite integral of f(x) = x^2 from x = 0 to x = 2.
Using Riemann Sum with 4 intervals:
- Δx = (2-0)/4 = 0.5
- f(x_1) = f(0) = 0, f(x_2) = f(0.5) = 0.25, f(x_3) = f(1) = 1, f(x_4) = f(1.5) = 2.25, f(x_5) = f(2) = 4
- Riemann Sum = f(x_1)Δx + f(x_2)Δx + f(x_3)Δx + f(x_4)Δx + f(x_5)Δx = (0)(0.5) + (0.25)(0.5) + (1)(0.5) + (2.25)(0.5) + (4)(0.5) = 4.125
As the number of intervals approaches infinity, the Riemann Sum converges to the definite integral.

Definite Integral and Area Under the Curve

Definite integral can be used to find the exact area under a curve.
The area under the curve y = f(x) between x = a and x = b is given by ∫(a to b) f(x) dx. Example:
Find the exact area under the curve y = x^2 between x = 2 and x = 4.
The definite integral is given by ∫(2 to 4) x^2 dx.
Using the power rule, we have ∫(2 to 4) x^2 dx = (1/3)x^3 | from 2 to 4
Evaluating the integral, we get [(1/3)(4)^3] - [(1/3)(2)^3] = 64/3 - 8/3 = 56/3 square units.

Definite Integral as a Measure of Accumulation

Definite integrals can be used to measure accumulation of quantities over a certain interval.
When the integrand represents a rate or density, the definite integral gives the total accumulated quantity.
Example: If a graph represents the velocity of a moving object over time, the definite integral of the velocity function over a certain time interval gives the total displacement of the object during that interval.

Definite Integral and Average Value

The average value of a function f(x) over an interval [a, b] is given by (1/(b-a)) ∫(a to b) f(x) dx.
It represents the height of a horizontal line that divides the area under the curve into two equal parts. Example:
Find the average value of f(x) = x^2 over the interval [0, 4].
The average value is given by (1/(4-0)) ∫(0 to 4) x^2 dx.
Using the power rule, we have ∫(0 to 4) x^2 dx = (1/3)x^3 | from 0 to 4
Evaluating the integral, we get [(1/3)(4)^3] - [(1/3)(0)^3] = 64/3
Therefore, the average value of f(x) = x^2 over [0, 4] is (1/(4-0))(64/3) = 16/3.

Definite Integral and Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if f(x) is continuous on an interval [a, b] and F(x) is an antiderivative of f(x), then ∫(a to b) f(x) dx = F(b) - F(a).
This theorem provides a link between indefinite and definite integrals. Example:
Let f(x) = 2x be continuous on [1, 3].
The antiderivative of f(x) is F(x) = x^2 + C.
Applying the Fundamental Theorem of Calculus, we have ∫(1 to 3) 2x dx = F(3) - F(1)
Evaluating F(x) at 3 and 1, we get (3^2 + C) - (1^2 + C) = 9 + C - 1 - C = 8.

Definite Integral and Symmetry

Definite integrals of symmetric functions over symmetric intervals yield zero.
If f(x) is an odd function, then ∫(-a to a) f(x) dx = 0.
If f(x) is an even function, then ∫(-a to a) f(x) dx = 2 ∫(0 to a) f(x) dx. Example:
Let f(x) = x^3 be an odd function.
The definite integral over the interval [-2, 2] is ∫(-2 to 2) x^3 dx = 0.

Definite Integral and Change of Variables

Change of variables is a technique used to simplify the evaluation of definite integrals.
It involves substituting a new variable in place of the original variable.
The limits of integration may also need to be adjusted accordingly. Example:
Find ∫(0 to 1) x^2 √(1 - x^3) dx using the change of variables u = 1 - x^3.
Differentiating both sides, we get du = -3x^2 dx.
Rearranging it, dx = -(1/(3x^2)) du.
Substituting in the original integral, we have ∫(0 to 1) x^2 √(1 - x^3) dx = ∫(1 to 0) -(1/(3x^2)) u^(1/2) du.
Adjusting the limits of integration, we get ∫(1 to 0) -(1/(3u^(2/3))) u^(1/2) du = -∫(0 to 1) (1/3)u^(-1/6) du.

Definite Integral and Integration by Parts

Integration by parts is a technique used to evaluate definite integrals involving the product of two functions.
It involves choosing one function as a derivative and the other as an antiderivative.
The formula for integration by parts is ∫(a to b) u dv = [u v]_(a to b) - ∫(a to b) v du. Example:
Find ∫(0 to 1) x e^x dx using integration by parts.
Let u = x and dv = e^x dx.
Differentiating u, we get du = dx.
Integrating dv, we get v = e^x.
Applying the integration by parts formula, we have ∫(0 to 1) x e^x dx = [x e^x]_(0 to 1) - ∫(0 to 1) e^x dx
Evaluating the integral and plugging in the limits, we get [1 e^1 - 0 e^0] - [e^x]_(0 to 1) = e - 1 - (e^1 - e^0) = e - 2.

Definite Integral and Partial Fractions

Partial fractions is a technique used to simplify complex rational functions.
It involves decomposing the rational function into simpler fractions.
The resulting fractions can then be integrated separately. Example:
Find ∫(0 to 2) (x^2 + 3x + 2)/(x^3 + 3x^2 + 2x) dx using partial fractions.
First, factorize the denominator: x(x+1)(x+2).
Write the given fraction as A/x + B/(x+1) + C/(x+2).
Find the values of A, B, and C by equating the numerators: x^2 + 3x + 2 = A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1).
Multiply through by the common denominator and solve for A, B, and C.
Once the partial fractions are found, integrate each term separately and then evaluate the integral.

Definite Integral and Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate definite integrals involving expressions with radicals.
It involves substituting the variable with trigonometric functions to simplify the integral.
The appropriate substitution depends on the form of the integrand. Example:
Find ∫(0 to √3) √(9 - x^2) dx using trigonometric substitution.
Let x = 3sinθ, then dx = 3cosθ dθ.
Substituting the given limits, we have ∫(0 to π/3) √(9 - (3sinθ)^2) (3cosθ) dθ.
Simplifying the integrand, we get ∫(0 to π/3) √(9 - 9sin^2θ) (3cosθ) dθ.
Using the Pythagorean identity sin^2θ + cos^2θ = 1, we can simplify the expression further.
Evaluate the integral using the substituted variable and substitute back to get the final answer.