Definite Integral - Definite Integral As A Limit Of Finite Sums Introduction

Definite Integral - Introduction

In calculus, the definite integral is used to find the total accumulation of a quantity over a given interval.
It is the limit of the sum of infinitesimally small quantities.
The definite integral is denoted by ∫.
It has a lower limit and an upper limit, specifying the interval of integration.
It represents the area under a curve between these limits.

Properties of Definite Integral

Linearity: ∫(af(x) + bg(x)) dx = a∫f(x) dx + b∫g(x) dx
Constant Rule: ∫c dx = cx + C
Additivity: ∫f(x) dx + ∫g(x) dx = ∫(f(x) + g(x)) dx
Substitution Rule: ∫f(g(x))g’(x) dx = ∫f(u) du
Bounds: ∫[a, b] f(x) dx = -∫[b, a]f(x) dx

Fundamental Theorem of Calculus - Part 1

Part 1 of the fundamental theorem of calculus states that if F(x) is the antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
This theorem helps us to evaluate definite integrals by finding the antiderivative of a given function. Example: Let f(x) = 3x^2. Find ∫[1, 3] 3x^2 dx. Solution: Using the fundamental theorem of calculus, we have: ∫[1, 3] 3x^2 dx = F(3) - F(1) = (x^3) [3, 1] = (3^3) - (1^3) = 27 - 1 = 26.

Fundamental Theorem of Calculus - Part 2

Part 2 of the fundamental theorem of calculus states that if f(x) is continuous on [a, b] and F(x) is any antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
This theorem allows us to evaluate definite integrals by finding the antiderivative of a given function. Equation: ∫[a, b] f(x) dx = F(b) - F(a)

Definite Integral - Area Interpretation

The definite integral represents the area under a curve between the limits of integration.
When the function is positive, the area under the curve is positive.
When the function is negative, the area under the curve is negative.
The net area is the algebraic sum of the positive and negative areas. Examples:

Find the area under the curve y = x^2 between x = 1 and x = 3. Solution: ∫[1, 3] x^2 dx = (x^3/3) [3, 1] = (3^3/3) - (1^3/3) = 9 - 1/3 = 8 2/3.

Find the area under the curve y = -x^2 between x = -2 and x = 2. Solution: ∫[-2, 2] -x^2 dx = -(x^3/3) [-2, 2] = -(2^3/3) - (-2^3/3) = -8/3 + 8/3 = 0.

Definite Integral - Riemann Sum

The definite integral can be evaluated using Riemann sums.
Riemann sums divide the interval of integration into small subintervals.
The sum of areas of rectangles formed by these subintervals approximates the area under the curve.
As the number of subintervals approaches infinity, the Riemann sum approaches the exact value of the definite integral. Equation: ∫[a, b] f(x) dx = lim(n→∞) Σ f(xi)Δx Where xi are the sample points and Δx is the width of each subinterval. Example: Approximate ∫[0, 4] x^2 dx using a Riemann sum with 4 subintervals. Solution: Δx = (4 - 0)/4 = 1 xi = 0, 1, 2, 3 The Riemann sum is: Σ f(xi)Δx = (0^2)(1) + (1^2)(1) + (2^2)(1) + (3^2)(1) = 0 + 1 + 4 + 9 = 14.

Definite Integral - Definite Integral as a Limit of Finite Sums (Introduction)

The definite integral can be defined as the limit of a Riemann sum as the number of subintervals approaches infinity.
The Riemann sum is an approximation of the area under the curve using rectangles.
As the width of each rectangle approaches zero, the Riemann sum approaches the exact value of the definite integral.
The definite integral represents the total accumulation of a quantity over a given interval.
It can be used to find areas, volumes, and average values of functions.

Definite Integral - Riemann Sum (Equation)

The Riemann sum is calculated using the formula: ∫[a, b] f(x) dx = lim (n→∞) Σ f(xi)Δx where xi are the sample points and Δx is the width of each subinterval.
The Riemann sum is the sum of the areas of the rectangles formed by the subintervals.
By taking the limit as the number of subintervals approaches infinity, we obtain the exact value of the definite integral. Example: Approximate ∫[0, 4] x^2 dx using a Riemann sum with 4 subintervals. Solution: Δx = (4 - 0)/4 = 1 xi = 0, 1, 2, 3 The Riemann sum is: Σ f(xi)Δx = (0^2)(1) + (1^2)(1) + (2^2)(1) + (3^2)(1) = 0 + 1 + 4 + 9 = 14.

Definite Integral - Properties

Linearity: ∫(af(x) + bg(x)) dx = a∫f(x) dx + b∫g(x) dx
Constant Rule: ∫c dx = cx + C
Additivity: ∫f(x) dx + ∫g(x) dx = ∫(f(x) + g(x)) dx
Substitution Rule: ∫f(g(x))g’(x) dx = ∫f(u) du
Bounds: ∫[a, b] f(x) dx = -∫[b, a]f(x) dx These properties help simplify the evaluation of definite integrals and perform operations on them.

Definite Integral - Geometric Interpretation

The definite integral represents the area under a curve between the limits of integration.
The area can be positive or negative depending on the function and the limits.
The net area is the algebraic sum of the positive and negative areas.
When the function is positive, the area under the curve is positive.
When the function is negative, the area under the curve is negative. Examples:

Find the area under the curve y = x^2 between x = 1 and x = 3. Solution: ∫[1, 3] x^2 dx = (x^3/3) [3, 1] = (3^3/3) - (1^3/3) = 9 - 1/3 = 8 2/3.

Find the area under the curve y = -x^2 between x = -2 and x = 2. Solution: ∫[-2, 2] -x^2 dx = -(x^3/3) [-2, 2] = -(2^3/3) - (-2^3/3) = -8/3 + 8/3 = 0.

Fundamental Theorem of Calculus - Part 1

Part 1 of the Fundamental Theorem of Calculus states that if F(x) is the antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
This theorem allows us to evaluate definite integrals by finding the antiderivative of the given function. Example: Let f(x) = 3x^2. Find ∫[1, 3] 3x^2 dx. Solution: Using the Fundamental Theorem of Calculus, we have: ∫[1, 3] 3x^2 dx = F(3) - F(1) = (x^3) [3, 1] = (3^3) - (1^3) = 27 - 1 = 26.

Fundamental Theorem of Calculus - Part 2

Part 2 of the Fundamental Theorem of Calculus states that if f(x) is continuous on [a, b] and F(x) is any antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
This theorem allows us to evaluate definite integrals by finding the antiderivative of the given function. Equation: ∫[a, b] f(x) dx = F(b) - F(a) This theorem provides a powerful method to evaluate definite integrals.

Definite Integral - Applications

The definite integral has various applications in mathematics and real-world problems.
It can be used to find areas, volumes, and average values of functions.
It is used in physics to calculate work, energy, and displacement.
It is used in economics to calculate total revenue, profit, and cost.
It is used in probability to calculate probabilities and expected values. The ability to evaluate definite integrals is essential in many fields and disciplines.

Definite Integral - Definite Integral as a Limit of Finite Sums (Introduction)

The definite integral can be defined as the limit of a Riemann sum as the number of subintervals approaches infinity.
The Riemann sum is an approximation of the area under the curve using rectangles.
As the width of each rectangle approaches zero, the Riemann sum approaches the exact value of the definite integral.
The definite integral represents the total accumulation of a quantity over a given interval.
It can be used to find areas, volumes, and average values of functions.

Definite Integral - Riemann Sum (Equation)

The Riemann sum is calculated using the formula: ∫[a, b] f(x) dx = lim (n→∞) Σ f(xi)Δx where xi are the sample points and Δx is the width of each subinterval.
The Riemann sum is the sum of the areas of the rectangles formed by the subintervals.
By taking the limit as the number of subintervals approaches infinity, we obtain the exact value of the definite integral. Example: Approximate ∫[0, 4] x^2 dx using a Riemann sum with 4 subintervals. Solution: Δx = (4 - 0)/4 = 1 xi = 0, 1, 2, 3 The Riemann sum is: Σ f(xi)Δx = (0^2)(1) + (1^2)(1) + (2^2)(1) + (3^2)(1) = 0 + 1 + 4 + 9 = 14.

Definite Integral - Properties

Linearity: ∫(af(x) + bg(x)) dx = a∫f(x) dx + b∫g(x) dx
Constant Rule: ∫c dx = cx + C
Additivity: ∫f(x) dx + ∫g(x) dx = ∫(f(x) + g(x)) dx
Substitution Rule: ∫f(g(x))g’(x) dx = ∫f(u) du
Bounds: ∫[a, b] f(x) dx = -∫[b, a]f(x) dx These properties help simplify the evaluation of definite integrals and perform operations on them. Example: Evaluate ∫[0, 1] (2x^3 + 3x^2 - 4x + 1) dx. Solution: Using the linearity property, we can split the integral into four separate integrals: ∫[0, 1] (2x^3 + 3x^2 - 4x + 1) dx = ∫[0, 1] 2x^3 dx + ∫[0, 1] 3x^2 dx - ∫[0, 1] 4x dx + ∫[0, 1] 1 dx = (2/4)x^4 + (3/3)x^3 - (4/2)x^2 + (1)x [0, 1] = (1/2)x^4 + x^3 - 2x^2 + x [0, 1] = (1/2)(1^4) + (1^3) - 2(1^2) + (1) - ((1/2)(0^4) + (0^3) - 2(0^2) + (0)) = 1/2 + 1 - 2 + 1 = 1/2.

Definite Integral - Geometric Interpretation

The definite integral represents the area under a curve between the limits of integration.
The area can be positive or negative depending on the function and the limits.
The net area is the algebraic sum of the positive and negative areas.
When the function is positive, the area under the curve is positive.
When the function is negative, the area under the curve is negative. Examples:

Find the area under the curve y = x^2 between x = 1 and x = 3. Solution: ∫[1, 3] x^2 dx = (x^3/3) [3, 1] = (3^3/3) - (1^3/3) = 9 - 1/3 = 8 2/3.

Find the area under the curve y = -x^2 between x = -2 and x = 2. Solution: ∫[-2, 2] -x^2 dx = -(x^3/3) [-2, 2] = -(2^3/3) - (-2^3/3) = -8/3 + 8/3 = 0.

Fundamental Theorem of Calculus - Part 1

Part 1 of the Fundamental Theorem of Calculus states that if F(x) is the antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
This theorem allows us to evaluate definite integrals by finding the antiderivative of the given function. Example: Let f(x) = 3x^2. Find ∫[1, 3] 3x^2 dx. Solution: Using the Fundamental Theorem of Calculus, we have: ∫[1, 3] 3x^2 dx = F(3) - F(1) = (x^3) [3, 1] = (3^3) - (1^3) = 27 - 1 = 26.

Fundamental Theorem of Calculus - Part 2

Part 2 of the Fundamental Theorem of Calculus states that if f(x) is continuous on [a, b] and F(x) is any antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
This theorem allows us to evaluate definite integrals by finding the antiderivative of the given function. Equation: ∫[a, b] f(x) dx = F(b) - F(a) This theorem provides a powerful method to evaluate definite integrals.

Definite Integral - Applications

The definite integral has various applications in mathematics and real-world problems.
It can be used to find areas, volumes, and average values of functions.
It is used in physics to calculate work, energy, and displacement.
It is used in economics to calculate total revenue, profit, and cost.
It is used in probability to calculate probabilities and expected values. The ability to evaluate definite integrals is essential in many fields and disciplines.