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Writing a Definite Integral In Exercises \(5 - 10\) , write a definite integral that represents the area of the region. (Do not evaluate the integral.) $$\begin{array} { l } { y _ { 1 } = x ^ { 2 } - 4 x + 3 } \\ { y _ { 2 } = - x ^ { 2 } + 2 x + 3 } \end{array}$$

Short Answer

Expert verified
The definite integral representing the area is \( \int_{0}^{3} |2x^2 - 6x| dx \)

Step by step solution

01

Identify intersections

The two curves intersect at points where \(y_{1} = y_{2}\). So, set \(x ^ { 2 } - 4 x + 3 = - x ^ { 2 } + 2 x + 3\) and solve for \(x\).
02

Solve for intersection points

Upon solving the equation, we get \(2x^2 - 6x = 0\). The solutions are \(x = 0\) and \(x = 3\).
03

Write the integral

The area between these two curves can be represented by the definite integral from 0 to 3 of the absolute difference of the functions. It can be written as \( \int_{0}^{3} |x^2 - 4x + 3 - (-x^2 + 2x + 3)| dx \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intersection Points
Finding intersection points of curves is crucial when determining the area between them. These points serve as the boundaries for our integration. To find where two curves intersect, we set their equations equal to each other. For this exercise, we set \(y_1 = y_2\), leading to the equation \(x^2 - 4x + 3 = -x^2 + 2x + 3\). This setup allows us to eliminate \(y\) from the equation, focusing on \(x\) alone.

Solving for \(x\), we bring all terms to one side and simplify: \(2x^2 - 6x = 0\). Factoring out \(2x\) gives us \(2x(x - 3) = 0\). Thus, \(x = 0\) and \(x = 3\) are our solutions. These solutions represent the x-coordinates where the two polynomial functions intersect. These points define the interval of our definite integral which helps in computing the area between the curves.
Area Between Curves
The area between two curves can be found by integrating the absolute difference of their functions over the interval of intersection points. This is because the integral calculates the accumulation of the vertical distances between the curves over a specified interval.

When given two functions \(y_1\) and \(y_2\), the absolute difference is expressed as \(|y_1 - y_2|\). This ensures we're getting the positive distance regardless of which function is on top within the interval. For this problem, the functions are \(y_1 = x^2 - 4x + 3\) and \(y_2 = -x^2 + 2x + 3\).
  • Determine the difference: \((x^2 - 4x + 3) - (-x^2 + 2x + 3) = 2x^2 - 6x\).
  • Set up the integral with limits derived from the intersection points: \(\int_{0}^{3} |2x^2 - 6x| \, dx\).
This integral calculates the area by summing up the small areas between the curves from \(x = 0\) to \(x = 3\), covering all the space between these boundaries.
Polynomial Functions
Polynomial functions, such as \(x^2 - 4x + 3\) and \(-x^2 + 2x + 3\) seen here, are expressions made up of variables and coefficients arranged in terms of powers. The leading exponent tells us the degree of the polynomial, which influences the shape of its graph. In our example, both functions are quadratic, being degree 2 polynomials, depicted as parabolas on a graph.

Quadratic functions follow a standard pattern: \(ax^2 + bx + c\). Each term plays a role:
  • The \(a\) coefficient affects the parabola's direction and width. A positive \(a\) opens upward, while a negative \(a\) opens downward.
  • The \(b\) term influences the parabola's position along the x-axis.
  • The \(c\) represents the y-intercept, where the function cuts through the y-axis.
Understanding polynomial structures helps in recognizing intersections and calculating areas, as having a firm grasp on how these functions behave allows us to predict where they might intersect and how they are positioned in the plane.

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Most popular questions from this chapter

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