Definite Integral

  • Definition: The definite integral of a function represents the area under the curve between two points on the x-axis.
  • We denote the definite integral as ∫(from a to b) f(x) dx.
  • The limits of integration, a and b, indicate the starting and ending points of integration.

Properties of Definite Integrals

  • Linearity: ∫(from a to b) [f(x) + g(x)] dx = ∫(from a to b) f(x) dx + ∫(from a to b) g(x) dx
  • Constant Multiplication: ∫(from a to b) c * f(x) dx = c * ∫(from a to b) f(x) dx
  • If a = b, then ∫(from a to a) f(x) dx = 0

Evaluation of Definite Integrals

  • The Fundamental Theorem of Calculus helps us evaluate definite integrals.
  • If F(x) is an antiderivative of f(x), then ∫(from a to b) f(x) dx = F(b) - F(a)
  • The value of the definite integral represents the net change in the antiderivative function between a and b.

Example: Evaluation of Definite Integral

Find ∫(from 1 to 4) 3x^2 dx Solution:

  • Let’s find the antiderivative of 3x^2: F(x) = x^3
  • Applying the Fundamental Theorem of Calculus, ∫(from 1 to 4) 3x^2 dx = F(4) - F(1)
  • = 4^3 - 1^3
  • = 64 - 1
  • = 63

Properties of Definite Integrals

  • If f(x) ≥ 0 on the interval [a, b], then the definite integral ∫(from a to b) f(x) dx represents the area under the curve.
  • If f(x) ≤ 0 on the interval [a, b], then ∫(from a to b) f(x) dx represents the negative of the area under the curve.
  • ∫(from a to b) -f(x) dx = -∫(from a to b) f(x) dx

Example: Evaluation of Definite Integral

Find ∫(from -2 to 2) |x| dx Solution:

  • The function |x| can be split into two separate cases:
    • For x ≥ 0, |x| = x
    • For x < 0, |x| = -x
  • Splitting the interval of integration, we get ∫(from -2 to 0) -x dx + ∫(from 0 to 2) x dx
  • Applying the Fundamental Theorem of Calculus, ∫(from -2 to 0) -x dx + ∫(from 0 to 2) x dx = -[x^2/2] from -2 to 0 + [x^2/2] from 0 to 2
  • = -[(0)^2/2 - (-2)^2/2] + [(2)^2/2 - (0)^2/2]
  • = -4 + 2
  • = -2

Definite Integral of a Constant Function

  • If f(x) is a constant function, then the definite integral ∫(from a to b) f(x) dx is equal to the product of the constant and the width of the interval.
  • ∫(from a to b) c dx = c * (b - a)

Example: Definite Integral of a Constant Function

Find ∫(from 1 to 4) 4 dx Solution:

  • Given the constant function f(x) = 4
  • Applying the property, ∫(from 1 to 4) 4 dx = 4 * (4 - 1)
  • = 4 * 3
  • = 12

Evaluation of Definite Integral using Antiderivatives

  • Sometimes, it is easier to evaluate definite integrals using antiderivatives.
  • We find the antiderivative of the integrand, substitute the limits of integration, and evaluate the antiderivative at those points.

Example: Evaluation of Definite Integral using Antiderivatives

Find ∫(from 0 to π) sin(x) dx Solution:

  • The antiderivative of sin(x) is -cos(x).
  • Applying the Fundamental Theorem of Calculus, ∫(from 0 to π) sin(x) dx = (-cos(π)) - (-cos(0))
  • = -(-1) - (-1)
  • = 1 - (-1)
  • = 2

Definite Integral - Integral Example using Antiderivative

  • Example:
    • Find ∫(from 1 to 3) (2x + 1) dx using antiderivatives.
  • Solution:
    • The antiderivative of 2x + 1 is x^2 + x.
    • Applying the Fundamental Theorem of Calculus, ∫(from 1 to 3) (2x + 1) dx = [(3)^2 + 3] - [(1)^2 + 1]
    • = (9 + 3) - (1 + 1)
    • = 12 - 2
    • = 10

Definite Integral as Net Change

  • The definite integral of a function can also represent the accumulation of a quantity over an interval.
  • If the function f(x) represents the rate at which a quantity is changing, then the definite integral ∫(from a to b) f(x) dx represents the total change in the quantity from a to b.

Example: Definite Integral as Net Change

  • Example:
    • The rate at which water is flowing in a pipe at time t is given by the function f(t) = 2t. Find the total amount of water that has flowed from t = 1 to t = 4.
  • Solution:
    • We need to find ∫(from 1 to 4) 2t dt.
    • Evaluating the integral, ∫(from 1 to 4) 2t dt = [t^2] from 1 to 4
    • = 4^2 - 1^2
    • = 16 - 1
    • = 15

The Definite Integral as Area Under the Curve

  • Integration can be thought of as the reverse process of differentiation.
  • The definite integral of a positive function represents the area under the curve between two points on the x-axis.

Example: Definite Integral as Area Under the Curve

  • Example:
    • Find the area bounded by the x-axis, the line y = 0, and the curve y = x^2 from x = -2 to x = 2.
  • Solution:
    • We need to find ∫(from -2 to 2) x^2 dx.
    • Evaluating the integral, ∫(from -2 to 2) x^2 dx = [x^3/3] from -2 to 2
    • = 2^3/3 - (-2)^3/3
    • = 8/3 + 8/3
    • = 16/3

Definite Integral of Piecewise Functions

  • Piecewise functions have different definitions for different intervals.
  • To find the definite integral of a piecewise function, we split the integral into separate parts based on the intervals and evaluate each part individually.

Example: Definite Integral of Piecewise Functions

  • Example:
    • Find ∫(from -2 to 2) f(x) dx, where
      • f(x) = 2x + 3 for -2 ≤ x < 0
      • f(x) = -2x + 3 for 0 ≤ x ≤2
  • Solution:
    • Splitting the integral, ∫(from -2 to 2) f(x) dx = ∫(from -2 to 0) (2x + 3) dx + ∫(from 0 to 2) (-2x + 3) dx
    • Integrating each part, = [(x^2 + 3x)] from -2 to 0 + [(-x^2 + 3x)] from 0 to 2
    • = [(0^2 + 3(0)) - ((-2)^2 + 3(-2))] + [((2)^2 + 3(2)) - (0^2 + 3(0))]
    • = (0 + 6) + (4 + 6)
    • = 6 + 10
    • = 16

Definite Integral with Symmetry

  • If a function is symmetric about the y-axis, then the definite integral over a symmetric interval is equal to 0.
  • This is because the positive and negative areas under the curve cancel each other out.

Example: Definite Integral with Symmetry

  • Example:
    • Find ∫(from -2 to 2) x^3 dx.
  • Solution:
    • Since f(x) = x^3 is an odd function, it is symmetric about the y-axis.
    • The definite integral ∫(from -2 to 2) x^3 dx is equal to 0.

Solving Definite Integrals Using Substitution

  • Sometimes, it is useful to use substitution to simplify the integrand and make it easier to evaluate the definite integral.
  • We choose a suitable substitution such that the integrand becomes simpler.
  • After evaluating the integral using the substituted variable, we transform the result back to the original variable.

Definite Integral and Area Between Curves

  • The definite integral can also be used to find the area between two curves.
  • To find the area between two curves, we first find the points of intersection.
  • We set the two equations equal to each other and solve for the x-values.
  • Then, we subtract the area under the bottom curve from the area under the top curve by evaluating the definite integral.

Example: Definite Integral for Area Between Curves

  • Example:
    • Find the area enclosed between the curves y = x^2 and y = 2x - 1.
  • Solution:
    • Setting the two equations equal to each other, x^2 = 2x - 1.
    • Rearranging the equation: x^2 - 2x + 1 = 0.
    • Solving for x, we find x = 1.
    • The area enclosed between the curves is given by ∫(from 1 to 2) (2x - 1 - x^2) dx.

Riemann Sums and Approximation of Definite Integral

  • Riemann sums are used to approximate the value of a definite integral.
  • The interval [a, b] is divided into n subintervals of equal width Δx = (b - a)/n.
  • A sample point is chosen from each subinterval, usually the left endpoint or right endpoint.
  • Riemann sum formula: ∑(from i = 1 to n) f(x_i) Δx.

Example: Riemann Sums and Approximation

  • Example:
    • Approximate the value of ∫(from 0 to 4) x^2 dx using 4 subintervals and left endpoints.
  • Solution:
    • Dividing the interval [0, 4] into 4 subintervals gives us Δx = (4 - 0)/4 = 1.
    • Using left endpoints, the Riemann sum is given by ∑(from i = 1 to 4) f(x_i) Δx, where x_i = (i-1)Δx.
    • Evaluating the sum: f(0)Δx + f(1)Δx + f(2)Δx + f(3)Δx = 0^2(1) + 1^2(1) + 2^2(1) + 3^2(1) = 0 + 1 + 4 + 9 = 14.

Indefinite Integral

  • The indefinite integral, or antiderivative, of a function f(x) represents the family of functions whose derivative is equal to f(x).
  • We denote the indefinite integral as ∫ f(x) dx.
  • The indefinite integral is the reverse process of differentiation.

Properties of Indefinite Integral

  • Linearity: ∫ (cf(x) + g(x)) dx = c ∫ f(x) dx + ∫ g(x) dx
  • Constant Multiplication: ∫ c f(x) dx = c ∫ f(x) dx
  • Power Rule: ∫ x^n dx = (x^(n+1))/(n+1) + C, where n ≠ -1

Example: Indefinite Integral using Power Rule

  • Example:
    • Find the indefinite integral of f(x) = 3x^2 + 2x + 1.
  • Solution:
    • Applying the power rule, the indefinite integral is given by ∫ (3x^2 + 2x + 1) dx = (3/3)x^3 + (2/2)x^2 + x + C = x^3 + x^2 + x + C.

Integration Techniques: Integration By Parts

  • Integration by parts is a technique used to evaluate the integral of a product of two functions.
  • The formula for integration by parts is: ∫ u dv = uv - ∫ v du, where u and v are functions of x.
  • We choose u and dv in such a way that the integral on the right-hand side is easier to evaluate than the original integral.

Example: Integration By Parts

  • Example:
    • Find the integral of f(x) = x*e^x dx.
  • Solution:
    • We choose u = x and dv = e^x dx.
    • Taking derivatives and integrating, we find du = dx and v = e^x.
    • Applying the formula for integration by parts, ∫ (xe^x) dx = xe^x - ∫ (e^x) dx.
    • Evaluating the integral, ∫ (e^x) dx = e^x.
    • Simplifying the expression, ∫ (xe^x) dx = xe^x - e^x + C.

Summary

  • Definite integral represents the area under the curve between two points on the x-axis.
  • Properties of definite integrals include linearity and constant multiplication.
  • Evaluation of definite integrals can be done using antiderivatives, substitution, or Riemann sums.
  • Indefinite integral, or antiderivative, represents the family of functions whose derivative is equal to the integrand.
  • Integration techniques such as integration by parts can be used to evaluate more complex integrals.