Problem 20 - Calculus of a Single Variable

Question

Finding the Area of a Region In Exercises \(17-30,\) sketch the region bounded by the graphs of the equations and find the area of the region. $$ y=-x^{2}+3 x+1, \quad y=-x+1 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the given equations is given by the result of the integral computed in Step 4.

Step 1: Sketch the Graphs

Draw the graphs of the two given equations. The first graph would be a parabola that opens downwards (-\(x^2\)) with its vertex at the point where \(x = 3/2\). The second graph is a straight line with a negative slope of -1 and y-intercept of +1.

Step 2: Find the Intersection Points

Set the two equations equal to \(y\) to find where they intersect, that is solve the equation \(-x^2 + 3x + 1 = -x + 1\) for \(x\). This will yield two solutions, which will represent the x-coordinates of the points where the two graphs intersect.

Step 3: Integrate to Find Area

Next, to find the area between the two curves, set up the integral from the lower x-coordinate to the upper x-coordinate of the absolute difference of the two functions, i.e., \(\int_{x_1}^{x_2} |-x^2 + 3x + 1 - (-x + 1)| dx\).

Step 4: Compute the Integral

Perform the integration and calculate the result. This result will give the area of the region bounded by the graphs of the two equations.

Key Concepts

Understanding Parabolas

A parabola is a U-shaped curve that can either open upwards or downwards. In this exercise, our parabola is given by the equation \( y = -x^2 + 3x + 1 \). Here, the negative coefficient of \( x^2 \) indicates that the parabola opens downwards. The vertex of a parabola is the highest or lowest point on the graph, depending on its direction. For this specific parabola, you calculate the vertex by using the formula \( -\frac{b}{2a} \) for the \( x \)-coordinate, referring to the general equation \( ax^2 + bx + c \). This places the vertex at \( x = \frac{3}{2} \).

• The vertex is a crucial point as it helps define the shape of the parabola.
• Knowing whether a parabola opens upward or downward assists in determining the region of interest when finding areas between curves.

Finding Intersection Points

Intersection points occur where two graphs meet or cross each other. To find the intersection points between the parabola \( y = -x^2 + 3x + 1 \) and the line \( y = -x + 1 \), set their equations equal: \(-x^2 + 3x + 1 = -x + 1\). This leads to solving a quadratic equation in order to find the \( x \)-coordinates where they intersect.

• Solve the equation to get the intersection points' \( x \)-coordinates.
• These coordinates indicate where to evaluate and set up the integration limits for finding the area between the curves.

By solving, you'll find two solutions, which specify the x-limits for integration, crucial for determining the enclosed area.

Introduction to Definite Integral

The definite integral is a fundamental concept used to find the area under a curve within a specific interval. To determine the area between two curves, calculate the integral of the difference between the functions. In this case, find the integral of \( |-x^2 + 3x + 1 - (-x + 1)| \) from the lower to the upper intersection x-values.

• Set up the definite integral using the obtained intersection points as your limits of integration.
• Take the absolute difference to ensure you are only summing positive areas.

By solving this integral, you effectively determine the exact region encompassed by the two curves.

Graph Sketching and Visualization

Graph sketching involves plotting functions on the coordinate plane to visually represent them. This visual guide helps understand the relationships between different functions such as intersections and areas between them.

• For the parabola \( y = -x^2 + 3x + 1 \), note the vertex and direction before sketching.
• The line \( y = -x + 1 \) can be sketched using its slope and y-intercept by identifying where it crosses the y-axis and its downward slope.
• Visualizing the intersection helps in appropriately setting the limits for integration.

By sketching both graphs, you gain clarity on the positioning and area occupied by the intersecting region which confirms the calculation done through definite integration.