Definite Integral - Area bounded between y^2=2x and y=4x^2

  • In this lesson, we will explore the concept of definite integral and apply it to find the area bounded between two curves.
  • The curves we will be working with are:
    • y^2 = 2x
    • y = 4x^2

Introduction to Definite Integral

  • The definite integral is a mathematical tool used to calculate the area under a curve between two points on the x-axis.
  • It is denoted by the symbol ∫ and has the following notation: ∫f(x)dx, where f(x) is the function and dx is the infinitesimal change in x.
  • The definite integral calculates the net area between the function and the x-axis within the specified interval.

Computing Definite Integral

  • To compute the definite integral, we need to determine the upper and lower limits of integration.
  • In our case, the limits will be the x-values corresponding to the points of intersection between the two curves.
  • Let’s denote the upper limit as b and the lower limit as a.

Example: Finding the Limits of Integration

  • To find the limits of integration, we need to set the two equations equal to each other.
  • y^2 = 2x
  • y = 4x^2
  • Setting them equal and solving for x, we get: 4x^2 = 2x
  • Rearranging the equation gives us: 2x^2 - x = 0
  • Factoring out x, we have: x(2x - 1) = 0
  • Solving for x gives us two values: x = 0 and x = 0.5

Example: Determining the Upper and Lower Limits

  • To determine the upper and lower limits, we evaluate the x-values at each curve.
  • For y^2 = 2x, at x = 0, y = √(2 * 0) = 0
  • For y^2 = 2x, at x = 0.5, y = √(2 * 0.5) = 1
  • For y = 4x^2, at x = 0, y = 4 * 0^2 = 0
  • For y = 4x^2, at x = 0.5, y = 4 * 0.5^2 = 1

Example: Graphical Representation

  • Let’s plot the two curves on a graph to visualize the area we need to find:
    • Curve 1: y^2 = 2x
    • Curve 2: y = 4x^2

Graphical Representation

Insert a graph here showing the curves y^2 = 2x and y = 4x^2

Area Bounded by the Curves

  • To find the area bounded by the curves, we need to compute the definite integral of the difference between the two equations.
  • The formula for computing area with definite integrals is: Area = ∫ (upper limit, lower limit) f(x) dx

Area Formula

  • The formula to compute the area between two curves can be written as: Area Formula
  • In our case, f(x) = √(2x) and g(x) = 4x^2

Applying the Area Formula

  • Now that we have the formula, let’s apply it to find the area bounded by the curves:
  • Area = ∫ (0, 0.5) |√(2x) - 4x^2| dx
  • We will evaluate the definite integral to find the area between the two curves.

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Evaluating the Definite Integral

  • We need to evaluate the definite integral to find the area between the two curves:
  • ∫ (0, 0.5) |√(2x) - 4x^2| dx
  • To do this, we will split the integral into two parts based on the points of intersection.
  • Area = ∫ (0, 0.5) (√(2x) - 4x^2) dx + ∫ (0.5, 0.5) (4x^2 - √(2x)) dx

Evaluating the First Integral

  • Let’s evaluate the first part of the integral:
  • ∫ (0, 0.5) (√(2x) - 4x^2) dx
  • This integral can be split into two separate integrals:
  1. ∫ (0, 0.5) √(2x) dx
  1. ∫ (0, 0.5) 4x^2 dx
  • We can use the power rule for integration to find the antiderivatives of each term.

Evaluating the First Integral (contd.)

  1. ∫ (0, 0.5) √(2x) dx:
    • Using the power rule, we get: (2/3) * (2x)^(3/2) | (0, 0.5)
    • Plugging in the limits, we get: (2/3) * (2(0.5))^(3/2) - (2/3) * (2(0))^(3/2)
    • Simplifying gives: (2/3) * √2 - 0
  1. ∫ (0, 0.5) 4x^2 dx:
    • Using the power rule, we get: (4/3) * x^3 | (0, 0.5)
    • Plugging in the limits, we get: (4/3) * (0.5)^3 - (4/3) * (0)^3
    • Simplifying gives: (4/3) * 0.125 - 0

Evaluating the First Integral (contd.)

  • Now, let’s sum up the two integrals:
  1. ∫ (0, 0.5) √(2x) dx = (2/3) * √2
  1. ∫ (0, 0.5) 4x^2 dx = (4/3) * 0.125
  • Adding the two values together, we get: (2/3) * √2 + (4/3) * 0.125

Evaluating the Second Integral

  • Let’s evaluate the second part of the integral:
  • ∫ (0.5, 0.5) (4x^2 - √(2x)) dx
  • This integral can be split into two separate integrals:
  1. ∫ (0.5, 0.5) 4x^2 dx
  1. ∫ (0.5, 0.5) √(2x) dx
  • We can use the power rule for integration to find the antiderivatives of each term.

Evaluating the Second Integral (contd.)

  1. ∫ (0.5, 0.5) 4x^2 dx:
    • Using the power rule, we get: (4/3) * x^3 | (0.5, 0.5)
    • Plugging in the limits, we get: (4/3) * (0.5)^3 - (4/3) * (0.5)^3
    • Simplifying gives: (4/3) * 0 - (4/3) * 0
  1. ∫ (0.5, 0.5) √(2x) dx:
    • Using the power rule, we get: (2/3) * (2x)^(3/2) | (0.5, 0.5)
    • Plugging in the limits, we get: (2/3) * (2(0.5))^(3/2) - (2/3) * (2(0.5))^(3/2)
    • Simplifying gives: (2/3) * √2 - (2/3) * √2

Evaluating the Second Integral (contd.)

  • Now, let’s sum up the two integrals:
  1. ∫ (0.5, 0.5) 4x^2 dx = (4/3) * 0 - (4/3) * 0 = 0
  1. ∫ (0.5, 0.5) √(2x) dx = (2/3) * √2 - (2/3) * √2 = 0
  • Adding the two values together, we get:

0 + 0 = 0

Final Calculation

  • To find the total area, we add the results from the first and second integrals: (2/3) * √2 + (4/3) * 0.125 + 0 + 0 = (2/3) * √2 + (4/3) * 0.125
  • Simplifying this expression will give us the final result for the area bounded between the two curves.

Final Result

  • After simplifying, we get the final result for the area bounded between the two curves:
  • Area = (2/3) * √2 + (4/3) * 0.125
  • This is the exact value of the area and cannot be further simplified.

Summary

  • In this lesson, we learned about definite integrals and how to apply them to find the area bounded between two curves.
  • We explored the example of finding the area bounded between the curves y^2 = 2x and y = 4x^2.
  • We evaluated the definite integral and found the exact value of the area.
  • Definite integrals are a powerful tool in mathematics and have various applications in physics, engineering, and other fields.

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Application of Definite Integrals

  • Definite integrals have many applications in calculus.
  • They are used to calculate areas, volumes, and even solve real-world problems.
  • Some common applications include finding the area between curves, calculating volumes of solids of revolution, and determining average values of functions.

Example: Finding Volume of Revolution

  • Let’s consider the function f(x) = x^2 on the interval [0, 3].
  • If we were to revolve this curve around the x-axis, we can use definite integrals to find the volume of the resulting solid.
  • The formula for finding the volume of revolution is: V = π∫(0, 3) f(x)^2 dx

Example: Finding Average Value of a Function

  • Definite integrals can also help us determine the average value of a function on a given interval.
  • The formula to find the average value of a function f(x) on the interval [a, b] is:
    • Average value = (1 / (b - a)) ∫(a, b) f(x) dx

Example: Average Value of a Function (contd.)

  • Let’s consider the function f(x) = x^2 on the interval [0, 2].
  • To find the average value of this function, we can use the formula:
    • Average value = (1 / (2 - 0)) ∫(0, 2) x^2 dx
  • Evaluating this definite integral will give us the average value of the function on the given interval.

Properties of Definite Integrals

  • Definite integrals have several properties that make them useful in mathematical calculations:
  1. Linearity: ∫(a, b) (f(x) + g(x)) dx = ∫(a, b) f(x) dx + ∫(a, b) g(x) dx
  1. Constant Multiple: ∫(a, b) c * f(x) dx = c * ∫(a, b) f(x) dx
  1. Bounds Reversal: ∫(a, b) f(x) dx = -∫(b, a) f(x) dx

Definite Integral as Antiderivative

  • The Fundamental Theorem of Calculus connects definite integrals and antiderivatives:
  • 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 compute definite integrals by finding an antiderivative of the function and evaluating it at the limits.

Improper Definite Integrals

  • Sometimes, definite integrals may not have finite values if the function has vertical asymptotes or infinite discontinuities within the interval.
  • In such cases, we have to consider improper definite integrals.
  • Improper integrals can be evaluated as limits, with the limits of integration approaching the problematic points.

Example: Improper Definite Integral

  • Let’s consider the function f(x) = 1/x on the interval [1, ∞].
  • This function has an infinite discontinuity at x = 0.
  • To evaluate the improper definite integral, we take the limit as the upper bound approaches infinity:
    • ∫(1, ∞) 1/x dx = lim (b→∞) ∫(1, b) 1/x dx

Example: Improper Definite Integral (contd.)

  • Evaluating this improper definite integral involves finding the antiderivative of 1/x, which is ln|x|.
    • lim (b→∞) ∫(1, b) 1/x dx = lim (b→∞) ln|b| - ln|1|
  • Taking the limit as b approaches infinity, we get:
    • lim (b→∞) ln|b| - ln|1| = ∞ - 0 = ∞
  • Therefore, the value of the improper definite integral is infinity.

Conclusion

  • Definite integrals are fundamental tools in calculus for finding areas, volumes, and solving various mathematical problems.
  • They have important properties and connections with antiderivatives.
  • Improper definite integrals are used when the standard rules and bounds do not apply.
  • Understanding definite integrals is crucial for performing advanced calculus calculations and real-world applications.