Slide 1

  • Topic: Definite Integral - Area bounded between two curves

Slide 2

  • The definite integral represents the “signed” area bounded between the x-axis and the curves of the function within a given interval.
  • It is denoted as:
    • ∫[a, b] f(x) dx
    • Where ‘a’ and ‘b’ are the limits of integration, and ‘f(x)’ is the function.

Slide 3

  • To find the area bounded between two curves, we subtract the area under one curve from the area under the other curve.
  • This can be expressed as:
    • Area = ∫[a, b] |f(x) - g(x)| dx
    • Where ‘f(x)’ and ‘g(x)’ are the two functions.

Slide 4

  • Steps to find the area bounded between two curves:
    1. Determine the points of intersection between the two curves.
    2. Identify which curve is above the other within the given interval.
    3. Set up the definite integral as: ∫[a, b] |f(x) - g(x)| dx.
    4. Evaluate the integral to find the area.

Slide 5

  • Example 1:
    • Find the area bounded between the curves y = x^2 and y = 2x - 1.
    • Determine the points of intersection:
      • x^2 = 2x - 1.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = 2x - 1 is above y = x^2 in the given interval.
    • Set up the definite integral:
      • Area = ∫[a, b] (2x - 1 - x^2) dx.

Slide 6

  • Example 2:
    • Find the area bounded between the curves y = sin(x) and y = cos(x) in the interval [0, π/2].
    • Determine the points of intersection:
      • sin(x) = cos(x).
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = cos(x) is above y = sin(x) in the given interval.
    • Set up the definite integral:
      • Area = ∫[0, π/2] (cos(x) - sin(x)) dx.

Slide 7

  • Example 3:
    • Find the area bounded between the curves y = x^3 and y = x^2 in the interval [0, 1].
    • Determine the points of intersection:
      • x^3 = x^2.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = x^3 is above y = x^2 in the given interval.
    • Set up the definite integral:
      • Area = ∫[0, 1] (x^3 - x^2) dx.

Slide 8

  • Example 4:
    • Find the area bounded between the curves y = e^x and y = x^2 in the interval [-1, 2].
    • Determine the points of intersection:
      • e^x = x^2.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = x^2 is above y = e^x in the given interval.
    • Set up the definite integral:
      • Area = ∫[-1, 2] (x^2 - e^x) dx.

Slide 9

  • Example 5:
    • Find the area bounded between the curves y = ln(x) and y = x in the interval [1, e].
    • Determine the points of intersection:
      • ln(x) = x.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = x is above y = ln(x) in the given interval.
    • Set up the definite integral:
      • Area = ∫[1, e] (x - ln(x)) dx.

Slide 10

  • Recap:
    • The definite integral can be used to find the area bounded between two curves.
    • Steps to find the area:
      1. Determine the points of intersection.
      2. Identify which curve is above the other.
      3. Set up the definite integral.
      4. Evaluate the integral to find the area.

Slide 11

  • Example 6:
    • Find the area bounded between the curves y = x^3 and y = 2x in the interval [0, 2].
    • Determine the points of intersection:
      • x^3 = 2x.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = 2x is above y = x^3 in the given interval.
    • Set up the definite integral:
      • Area = ∫[0, 2] (2x - x^3) dx.

Slide 12

  • Example 7:
    • Find the area bounded between the curves y = 4 - 3x and y = -3x^2 + 2x + 3 in the interval [-1, 2].
    • Determine the points of intersection:
      • 4 - 3x = -3x^2 + 2x + 3.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = 4 - 3x is below y = -3x^2 + 2x + 3 in the given interval.
    • Set up the definite integral:
      • Area = ∫[-1, 2] ((-3x^2 + 2x + 3) - (4 - 3x)) dx.

Slide 13

  • Example 8:
    • Find the area bounded between the curves y = e^x and y = 2^x in the interval [-1, 1].
    • Determine the points of intersection:
      • e^x = 2^x.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = e^x is above y = 2^x in the given interval.
    • Set up the definite integral:
      • Area = ∫[-1, 1] (e^x - 2^x) dx.

Slide 14

  • Example 9:
    • Find the area bounded between the curves y = 1/x and y = 1/x^2 in the interval [1, 2].
    • Determine the points of intersection:
      • 1/x = 1/x^2.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = 1/x^2 is above y = 1/x in the given interval.
    • Set up the definite integral:
      • Area = ∫[1, 2] (1/x^2 - 1/x) dx.

Slide 15

  • Example 10:
    • Find the area bounded between the curves y = x^2 - 4x and y = -x^2 + 2x + 7 in the interval [1, 3].
    • Determine the points of intersection:
      • x^2 - 4x = -x^2 + 2x + 7.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = -x^2 + 2x + 7 is above y = x^2 - 4x in the given interval.
    • Set up the definite integral:
      • Area = ∫[1, 3] ((-x^2 + 2x + 7) - (x^2 - 4x)) dx.

Slide 16

  • Properties of definite integrals:
    1. Linearity: ∫[a, b] (cf(x) + dg(x)) dx = c∫[a, b] f(x) dx + d∫[a, b] g(x) dx
    2. Additivity: ∫[a, b] f(x) dx + ∫[b, c] f(x) dx = ∫[a, c] f(x) dx
    3. Change of limits: ∫[a, b] f(x) dx = -∫[b, a] f(x) dx

Slide 17

  • Fundamental Theorem of Calculus - Part 1:
    • If F(x) is an antiderivative of f(x) on the interval [a, b], then:
      • ∫[a, b] f(x) dx = F(b) - F(a)

Slide 18

  • Fundamental Theorem of Calculus - Part 2:
    • If F(x) is an antiderivative of f(x) on the interval [a, b], then:
      • d/dx ∫[a, x] f(t) dt = f(x)

Slide 19

  • Special cases of definite integrals:
    1. ∫[a, a] f(x) dx = 0
    2. ∫[a, b] 0 dx = 0
    3. ∫[a, a] c dx = 0
    4. ∫[a, a] f(x) dx = 0, if f(x) is continuous at ‘a’

Slide 20

  • Summary:
    • The definite integral can be found to calculate the area bounded between two curves.
    • Several examples were discussed to illustrate the process.
    • Properties and the Fundamental Theorem of Calculus were introduced.
    • Special cases of definite integrals were discussed.

Slide 21

  • Example 11:
    • Find the area bounded between the curves y = x^2 + 2x and y = 2x - 1 in the interval [-2, 1].
    • Determine the points of intersection:
      • x^2 + 2x = 2x - 1.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = 2x - 1 is above y = x^2 + 2x in the given interval.
    • Set up the definite integral:
      • Area = ∫[-2, 1] (2x - 1 - (x^2 + 2x)) dx.

Slide 22

  • Example 12:
    • Find the area bounded between the curves y = 3x^2 and y = 4 - x^2 in the interval [-2, 2].
    • Determine the points of intersection:
      • 3x^2 = 4 - x^2.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = 4 - x^2 is above y = 3x^2 in the given interval.
    • Set up the definite integral:
      • Area = ∫[-2, 2] (4 - x^2 - 3x^2) dx.

Slide 23

  • Example 13:
    • Find the area bounded between the curves y = e^x and y = e^(2x) in the interval [0, ln(2)].
    • Determine the points of intersection:
      • e^x = e^(2x).
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = e^x is below y = e^(2x) in the given interval.
    • Set up the definite integral:
      • Area = ∫[0, ln(2)] (e^(2x) - e^x) dx.

Slide 24

  • Example 14:
    • Find the area bounded between the curves y = x^3 - x^2 and y = 4 - x^3 in the interval [-1, 2].
    • Determine the points of intersection:
      • x^3 - x^2 = 4 - x^3.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = 4 - x^3 is below y = x^3 - x^2 in the given interval.
    • Set up the definite integral:
      • Area = ∫[-1, 2] (x^3 - x^2 - (4 - x^3)) dx.

Slide 25

  • Example 15:
    • Find the area bounded between the curves y = ln(x) and y = e^x in the interval [1, e].
    • Determine the points of intersection:
      • ln(x) = e^x.
    • Solve for ‘x’ to find the intersection points.
    • Identify the curves:
      • The curve y = ln(x) is below y = e^x in the given interval.
    • Set up the definite integral:
      • Area = ∫[1, e] (e^x - ln(x)) dx.

Slide 26

  • Properties of definite integrals (continued): 4. Constant multiple: ∫[a, b] c·f(x) dx = c∫[a, b] f(x) dx 5. Bounds: ∫[a, a] f(x) dx = 0 6. Bounds reversal: ∫[b, a] f(x) dx = -∫[a, b] f(x) dx

Slide 27

  • Properties of definite integrals (continued): 7. Symmetry: If f(x) is an even function, then ∫[-a, a] f(x) dx = 2∫[0, a] f(x) dx 8. Symmetry: If f(x) is an odd function, then ∫[-a, a] f(x) dx = 0 9. Linearity with limits: ∫[a, b] f(x) dx + ∫[b, c] f(x) dx = ∫[a, c] f(x) dx

Slide 28

  • Fundamental Theorem of Calculus - Part 3:
    • If f(x) is continuous on the closed interval [a, b] and F’(x) = f(x) for all x in the interval, then the definite integral of f(x) from a to b represents the net change of F(x) on that interval.
    • Mathematically, ∫[a, b] f(x) dx = F(b) - F(a)

Slide 29

  • Example 16:
    • Let F(x) = ∫[0, x] e^(2t) dt. Find F’(x).
    • By the Fundamental Theorem of Calculus - Part 1,
      • F’(x) = e^(2x)

Slide 30

  • Summary:
    • The definite integral can be used to calculate the area bounded between two curves.
    • Additional examples were provided for further practice.
    • Properties of definite integrals were discussed.
    • The Fundamental Theorem of Calculus was introduced and its applications were shown.