Definite Integral - Application of Definite Integral in Area of Simple Curve

Slide 1

  • In this lecture, we will discuss the application of definite integral in finding the area of a simple curve.

Slide 2

  • A simple curve is a continuous curve that can be defined by a single equation in a given interval.
  • For example, y = f(x) is a simple curve.

Slide 3

  • The definite integral is a mathematical tool used to find the area under a curve.
  • It is denoted by ∫ f(x) dx over the interval [a, b].

Slide 4

  • The definite integral of a function f(x) over the interval [a, b] gives the signed area between the curve and the x-axis.
  • The signed area has a positive value if the curve is above the x-axis, and a negative value if the curve is below the x-axis.

Slide 5

  • The area under a simple curve can be calculated using the definite integral as follows: A = ∫ f(x) dx over the interval [a, b]

Slide 6

  • To find the area under a simple curve, first determine the limits of integration, which are the x-values a and b.
  • Then, evaluate the definite integral using appropriate integration techniques.

Slide 7

  • Let’s consider an example to understand how to find the area under a simple curve.
  • Example: Find the area under the curve y = x^2 over the interval [0, 2].

Slide 8

  • Step 1: Determine the limits of integration, which are the x-values 0 and 2.
  • Step 2: Set up the definite integral: A = ∫ (x^2) dx over the interval [0, 2]

Slide 9

  • Step 3: Evaluate the definite integral: A = [(x^3)/3] from 0 to 2 A = [(2^3)/3] - [(0^3)/3] A = (8/3) - 0 A = 8/3

Slide 10

  • Therefore, the area under the curve y = x^2 over the interval [0, 2] is 8/3 square units.

(End of slides 1-10) Here are slides 11 to 20 for the topic “Definite Integral - Application of Definite Integral in Area of Simple Curve”:

Slide 11

  • The definite integral can also be used to find the area between two curves.
  • The area between two curves is given by the difference of their integral values.
  • For example, if we have two curves y = f(x) and y = g(x), the area between these curves is given by ∫ [f(x) - g(x)] dx over the interval [a, b].

Slide 12

  • Let’s consider an example to understand how to find the area between two curves.
  • Example: Find the area between the curves y = x^2 and y = x over the interval [0, 1].
  • Step 1: Determine the limits of integration, which are the x-values 0 and 1.
  • Step 2: Set up the definite integral: A = ∫ (x^2 - x) dx over the interval [0, 1]

Slide 13

  • Step 3: Evaluate the definite integral: A = [(x^3)/3 - (x^2)/2] from 0 to 1 A = [(1^3)/3 - (1^2)/2] - [(0^3)/3 - (0^2)/2] A = (1/3 - 1/2) - (0/3 - 0/2) A = (1/3 - 1/2) - 0 A = -1/6

Slide 14

  • Therefore, the area between the curves y = x^2 and y = x over the interval [0, 1] is -1/6 square units.
  • Note that the result is negative because the curve y = x^2 is below the curve y = x in the given interval.

Slide 15

  • The definite integral can also be used to find the area between a curve and the y-axis.
  • To find the area between a curve and the y-axis, we need to rearrange the integral to be in terms of y.
  • The area is given by ∫ x dy over the interval [c, d], where c and d are the y-values of the curve.

Slide 16

  • Let’s consider an example to find the area between a curve and the y-axis.
  • Example: Find the area between the curve x = y^2 and the y-axis over the interval [0, 2].
  • Step 1: Determine the limits of integration, which are the y-values 0 and 2.
  • Step 2: Rearrange the integral to be in terms of x: A = ∫ x dy over the interval [0, 2] A = ∫ √x dx over the interval [0, 2]

Slide 17

  • Step 3: Evaluate the definite integral: A = [(2/3) * x^(3/2)] from 0 to 2 A = [(2/3) * (2^(3/2))] - [(2/3) * (0^(3/2))] A = [(2/3) * (2^(3/2))] - 0 A = 2/3 * 2^(3/2)

Slide 18

  • Therefore, the area between the curve x = y^2 and the y-axis over the interval [0, 2] is 2/3 * 2^(3/2) square units.

Slide 19

  • In summary, the definite integral is a powerful tool used to find the area under a curve, the area between two curves, and the area between a curve and the y-axis.
  • It allows us to calculate complex areas with mathematical precision.

Slide 20

  • Understanding the concepts and applications of definite integrals is essential for various mathematical fields and real-life situations.
  • It provides a framework for solving problems related to area, volume, and other important calculations.

==(End of slides 11-20) ==

Definite Integral - Application of Definite Integral in Area of Simple Curve

Slide 21

  • Here are some important points to remember when using definite integrals to find the area of a simple curve:
    • The definite integral must be evaluated within the given limits of integration.
    • The integrand should represent the function that defines the curve.
    • The area can be positive or negative, depending on the position of the curve with respect to the x-axis.

Slide 22

  • Let’s consider another example to find the area under a simple curve.

  • Example: Find the area under the curve y = 2x + 1 over the interval [-1, 1].

  • Step 1: Determine the limits of integration, which are the x-values -1 and 1.

  • Step 2: Set up the definite integral: A = ∫ (2x + 1) dx over the interval [-1, 1]

  • Step 3: Evaluate the definite integral: A = [(x^2) + x] from -1 to 1 A = (1^2 + 1) - ((-1)^2 + (-1)) A = 2 - (-2) A = 4 square units

Slide 23

  • Therefore, the area under the curve y = 2x + 1 over the interval [-1, 1] is 4 square units.
  • This result is positive because the curve is always above the x-axis in the given interval.

Slide 24

  • The definite integral can also be used to find the area between two curves when they intersect.
  • When finding the area between two curves, it is important to determine the points of intersection to set the limits of integration.

Slide 25

  • Let’s consider an example to find the area between two curves.

  • Example: Find the area between the curves y = x^2 and y = 2x - 1.

  • Step 1: Determine the points of intersection by setting the equations equal to each other: x^2 = 2x - 1 x^2 - 2x + 1 = 0 (x - 1)^2 = 0 x = 1

  • Step 2: Determine the limits of integration, which are the x-values where the curves intersect. In this case, x = 1.

  • Step 3: Set up the definite integral: A = ∫ (x^2 - (2x - 1)) dx over the interval [0, 1]

Slide 26

  • Step 4: Evaluate the definite integral: A = [((x^3)/3) - ((x^2)/2) + x] from 0 to 1 A = [(1/3) - (1/2) + 1] - [(0/3) - (0/2) + 0] A = (1/3) - (1/2) + 1 A = 5/6 square units
  • Therefore, the area between the curves y = x^2 and y = 2x -1 over the interval [0, 1] is 5/6 square units.

Slide 27

  • The definite integral can be applied to various shapes and curves, not just simple curves.
  • It can be used to find the area of irregular shapes, volumes of solids, and even probabilities in statistics.

Slide 28

  • The concept of the definite integral and its applications are essential in various fields such as physics, engineering, economics, and computer science.
  • Understanding its principles and techniques will provide a strong foundation for further mathematical and scientific studies.

Slide 29

  • In conclusion, the definite integral is a powerful tool for finding the area under a curve and between curves.
  • It enables precise calculations of complex areas and provides valuable insights in various mathematical and real-life situations.

Slide 30

  • Thank you for attending this lecture on the application of definite integral in the area of a simple curve.
  • Practice solving different examples and exercises to further enhance your understanding of this topic.
  • Good luck with your studies and exams!