Horizontal area between curves | Applications of definite integrals | AP Calculus AB | Khan Academy

Khan Academy
1 Aug 201708:03
EducationalLearning
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TLDRThe video script presents a mathematical problem involving the intersection of two curves, where x is a function of y. The focus is on calculating the area between the curves by setting up a definite integral with respect to y, using the points of intersection as bounds. The solution involves finding the intersection points by equating the expressions of the two curves, simplifying the resulting quadratic equation, and then evaluating the antiderivative of the difference between the functions over the interval defined by the y-coordinates of the intersection points. The final answer is computed with careful attention to detail, resulting in a numerical value for the area.

Takeaways
  • 📈 The script discusses the process of finding the area between two curves where x is a function of y, not the traditional y as a function of x.
  • 🤔 The video encourages viewers to pause and attempt to solve the problem of finding the area between the curves themselves before revealing the method.
  • 🧩 The solution involves setting up a definite integral with respect to y, using the y-coordinates of the intersection points as the bounds of integration.
  • 📊 The lower bound of integration is denoted as y=1 and the upper bound as y=3, based on the intersection points of the two curves.
  • 🔍 Identifying the points of intersection is crucial, which is achieved by setting the two y-dependent x expressions equal to each other and simplifying the resulting quadratic equation.
  • 👌 The quadratic equation simplifies to -2y^2 + 2y + 12 = 0, which can be factored into -2(y-3)(y+2) = 0, giving the intersection points y=3 and y=-2.
  • 📝 The integral to be evaluated is from y=-2 to y=3 of (-y^2 + 3y + 11) - (y^2 + y - 1) dy, which simplifies to the integral of -2y^2 + 2y + 12 dy.
  • 🧮 The antiderivatives are calculated, resulting in -2y^3/3 + y^2 + 12y, which is evaluated at the bounds of the integral.
  • 🔢 The final calculation involves evaluating the antiderivative at the upper bound (y=3) and subtracting the evaluation at the lower bound (y=-2) to find the area.
  • 🎯 The area between the curves is found to be 47 - 5(1/3), which simplifies to 41 2/3 square units.
Q & A

  • What is the main topic of the video?

    -The main topic of the video is finding the area between two curves where x is a function of y, using definite integration with respect to y.

  • How does the instructor suggest finding the area between the curves?

    -The instructor suggests finding the area by integrating with respect to y, using a definite integral where the bounds are in terms of y, from the lower to the upper point of intersection.

  • What are the two functions given in the example?

    -The two functions given in the example are f(y) = -y^2 + 3y + 11 and g(y) = y^2 + y - 1.

  • How does the instructor describe the process of setting up the integral?

    -The instructor describes the process of setting up the integral by identifying the y-coordinates of the points of intersection as the bounds for the integral and then integrating the difference of the functions, (f(y) - g(y)), with respect to y.

  • What is the equation used to find the points of intersection?

    -The equation used to find the points of intersection is -y^2 + 3y + 11 = y^2 + y - 1. By setting the two expressions equal to each other and simplifying, the points of intersection are found.

  • How does the instructor simplify the equation to find the points of intersection?

    -The instructor simplifies the equation by subtracting y^2, y, and -1 from both sides, resulting in a quadratic equation -2y^2 + 2y + 12 = 0. This is then factored into -2(y - 3)(y + 2) = 0, giving the points of intersection at y = 3 and y = -2.

  • What is the definite integral set up to evaluate?

    -The definite integral set up to evaluate is from y = -2 to y = 3 of (-y^2 + 3y + 11) - (y^2 + y - 1) dy, which simplifies to the integral of (-2y^2 + 2y + 12) dy.

  • How does the instructor find the antiderivative of the integrand?

    -The instructor finds the antiderivative by applying the reverse power rule for integration. The antiderivative is -2y^3/3 + y^2 + 12y.

  • What are the steps to evaluate the definite integral?

    -The steps to evaluate the definite integral are to find the antiderivative, evaluate it at the upper bound (y = 3), and then subtract the value of the antiderivative at the lower bound (y = -2).

  • What is the final result of the area between the curves?

    -The final result of the area between the curves is 41 2/3 square units.

  • How does the video script illustrate the concept of integration?

    -The video script illustrates the concept of integration by using the example of finding the area between two curves. It explains the process of setting up the integral, finding the points of intersection, simplifying the integrand, finding the antiderivative, and evaluating the definite integral to find the area.

  • What is the significance of the points of intersection in this context?

    -The points of intersection are significant because they provide the bounds for the definite integral that is used to calculate the area between the curves. Without these points, the limits of integration would not be known.

Outlines
00:00
📚 Introduction to Integrating with Respect to y

The first paragraph introduces the concept of integrating with respect to y, highlighting the unusual setup where x is a function of y. The instructor presents two curves and explains the goal of finding the area between them. A key hint is provided, suggesting the use of a definite integral with bounds in terms of y. The lower and upper bounds of integration are identified as y1 and y2, corresponding to the y-coordinates of the points of intersection of the two curves. The concept of integrating as the sum of infinitely thin rectangles is introduced, with dy representing the height and the difference in x-values (f(y) - g(y)) representing the width.

05:00
🧮 Solving the Intersection Points and Evaluating the Integral

In the second paragraph, the process of finding the points of intersection is detailed, which is essential for setting up the integral. The instructor sets the two y expressions equal to each other to find the intersection points by solving a quadratic equation. After simplifying and factoring the equation, the intersection points are determined to be y = -2 and y = 3. The integral is then evaluated from y = -2 to y = 3, with the antiderivative of the integrand being calculated. The final steps involve evaluating the antiderivative at the bounds and computing the difference to find the area, resulting in a final answer of 41 2/3.

Mindmap
Keywords
💡Curves
Curves refer to the continuous lines that represent the graph of a function. In the video, the instructor has two curves graphed, which are mathematical representations of two different functions or relationships between variables x and y. The curves are essential to understanding the problem as they visually demonstrate the relationship and the area between them that needs to be calculated.
💡Functions
Functions are mathematical relationships between two variables, typically denoted as x and y, where each value of x corresponds to a specific value of y. In the video, the functions are represented by the equations of the curves. The main focus is on finding the area between these functions, which requires understanding how x is related to y in each equation.
💡Integration
Integration is a fundamental operation in calculus that has to do with finding the accumulated quantity under a curve. In the context of the video, integration with respect to y is used to calculate the area between the two curves. The process involves setting up a definite integral with bounds determined by the y-coordinates of the points of intersection of the curves.
💡Definite Integral
A definite integral represents the exact numerical value of the accumulated quantity under a curve and between two specified bounds. In the video, the definite integral is used to calculate the area between the two curves by summing up an infinite number of infinitely thin rectangles, with the height of each rectangle being dy and the width being the difference in x-values of the functions at each y-value.
💡Bounds
Bounds in the context of integration refer to the specified limits or intervals within which the integration is performed. In the video, the bounds are the y-coordinates of the points where the two curves intersect, which are used to define the interval for the definite integral.
💡Infinitely Thin Rectangles
The concept of infinitely thin rectangles is a visual way to understand the process of integration. It involves dividing the area under the curve into countless small rectangles with infinitesimal height (dy) and calculating the sum of these rectangles' areas to approximate the area under the curve. In the video, this concept is used to explain how the definite integral calculates the area between the two curves.
💡Antiderivative
The antiderivative, also known as the indefinite integral, is a function that represents the reverse process of differentiation. It is used in integration to find the original function whose derivative would match the function being integrated. In the video, the antiderivative is calculated to evaluate the definite integral and find the area between the curves.
💡Points of Intersection
Points of intersection refer to the specific coordinates where two or more curves meet on a graph. In the video, the points of intersection are crucial for determining the bounds of the definite integral and for setting up the equation that leads to finding the y-coordinates of these points.
💡Quadratic Equation
A quadratic equation is a polynomial equation of degree two, which typically takes the form of ax squared plus bx plus c equals zero. In the video, the process of finding the points of intersection involves setting up and solving a quadratic equation derived from equating the two functions of the curves.
💡Factoring
Factoring is the process of breaking down a polynomial into a product of other polynomials or factors. In the video, factoring is used to simplify the quadratic equation and find the values of y that satisfy it, which are the points of intersection of the two curves.
💡Area Calculation
Area calculation is the process of determining the amount of space enclosed within a shape. In the context of the video, the area calculation is performed by evaluating the definite integral, which sums up the areas of infinitely thin rectangles under the curves to find the enclosed area between them.
Highlights

The video discusses the concept of finding the area between two curves where x is a function of y, which is a different perspective from the usual y as a function of x.

The method of integration with respect to y is introduced as a way to calculate the area between the two curves.

The importance of identifying the points of intersection of the curves is emphasized to establish the bounds for the integral.

The process of setting up the integral involves taking the difference of the functions f(y) and g(y) over the interval defined by the y-coordinates of the intersection points.

The video demonstrates the algebraic manipulation of the given functions to find the points of intersection by setting them equal to each other and simplifying the resulting quadratic equation.

The solution of the quadratic equation is obtained through factoring, revealing the intersection points at y = -2 and y = 3.

The definite integral is set up from y = -2 to y = 3, incorporating the expressions for f(y) and g(y).

The calculation involves taking the antiderivative of the integrand and evaluating it at the bounds of the integral.

The process of evaluating the antiderivative at the upper bound (y = 3) and the lower bound (y = -2) is detailed, highlighting the steps of the calculation.

The final result of the area calculation is presented as 27, obtained by summing the results of the upper and lower bounds' evaluations.

The video provides a comprehensive walkthrough of the problem, including the mathematical reasoning and the steps to solve for the area between two curves.

The method showcased in the video can be applied to various problems involving the calculation of areas in the context of functions and integration.

The video emphasizes the practical application of integration in finding areas between curves, which is a fundamental concept in calculus.

The detailed explanation of the steps, from setting up the integral to evaluating it, makes the video a valuable resource for learning and understanding the process.

The video demonstrates the use of the reverse power rule and factoring in solving the integral, showcasing important techniques in algebra and calculus.

The final answer of 41 2/3 is derived after a thorough and methodical calculation, illustrating the precision required in mathematical problem-solving.

Transcripts

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