Problem 20 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=x^{2}, y=2-x, y=1$$

Step-by-Step Solution

Answer

The total area bounded by the curves is 3 square units.

Step 1: Understand the Region of Interest

To find the area between curves, first establish the region of interest. The curves given are: \( y = x^2 \), \( y = 2 - x \), and \( y = 1 \). Investigate where these curves intersect to determine the bounded region.

Step 2: Find Points of Intersection

Find the intersections of \( y = x^2 \) and \( y = 2 - x \) by equating: \ \( x^2 = 2 - x \) \ which simplifies to \( x^2 + x - 2 = 0 \). Factoring gives: \ \( (x - 1)(x + 2) = 0 \) \ Therefore, \( x = 1 \) and \( x = -2 \). \ For \( y = x^2 \) and \( y = 1 \), solve: \ \( x^2 = 1 \) \ which gives \( x = 1 \) and \( x = -1 \). \ For \( y = 2 - x \) and \( y = 1 \), solve: \ \( 2 - x = 1 \) \ which gives \( x = 1 \). \ The relevant points are at \( x = -2 \), \( x = -1 \), and \( x = 1 \).

Step 3: Determine Bounds and Divide the Region

The area to find is from \( x = -2 \) to \( x = 1 \). The segment from \( x = -2 \) to \( x = -1 \) is bounded by \( y = 2 - x \) (top) and \( y = x^2 \) (bottom). From \( x = -1 \) to \( x = 1 \), it's bounded at the top by \( y = 1 \) and at the bottom by \( y = x^2 \).

Step 4: Calculate the Area from x = -2 to x = -1

The area between \( y = 2 - x \) and \( y = x^2 \) from \( x = -2 \) to \( x = -1 \) is: \ \( \int_{-2}^{-1} ((2 - x) - x^2) \, dx \) \ Evaluating, \ \( \int_{-2}^{-1} (2 - x - x^2) \, dx = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{-1} \) \ Calculate: \ For \( x = -1: \) \( 2(-1) - \frac{(-1)^2}{2} - \frac{(-1)^3}{3} \) \ For \( x = -2: \) \( 2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} \) \ Evaluate and subtract to get the area.

Step 5: Calculate the Area from x = -1 to x = 1

The area between \( y = 1 \) and \( y = x^2 \) from \( x = -1 \) to \( x = 1 \) is: \ \( \int_{-1}^{1} (1 - x^2) \, dx \) \ Evaluate: \ \( \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} \) \ Calculate for \( x = 1 \) and \( x = -1 \), then subtract to find the area.

Step 6: Sum the Areas

Add the results from Steps 4 and 5 to find the total area.

Key Concepts

Points of Intersection

When dealing with the problem of finding the area between curves, the first step is always to pinpoint where these curves intersect. This set of intersection points helps define the exact region we're interested in, guiding the bounds for our calculations.
To find where curves intersect, set the equations equal to one another. For example, to find where the curves of equations \( y = x^2 \) and \( y = 2 - x \) intersect, solve \( x^2 = 2 - x \). This simplifies to \( x^2 + x - 2 = 0 \).
Use factoring techniques to solve this quadratic equation, which results in \( (x - 1)(x + 2) = 0 \). Thus, the points \( x = 1 \) and \( x = -2 \) are intersections. Repeating similar steps for other combinations of curves will reveal additional points of intersection, important for defining separate bounded regions for integration.

Definite Integral

The definite integral is a powerful tool for calculating the area under or between curves over a specified interval. Once the points of intersection are identified, they offer the bounds for this integral calculation.
For the problem at hand, consider the integral between the points \( x = -2 \) and \( x = -1 \) for the curves \( y = 2 - x \) (top) and \( y = x^2 \) (bottom). The definite integral is expressed as \( \int_{-2}^{-1} ((2 - x) - x^2) \, dx \). Evaluating this gives the exact area for that segment.
Another integral from \( x = -1 \) to \( x = 1 \) is needed for curves \( y = 1 \) (top) and \( y = x^2 \) (bottom), calculated as \( \int_{-1}^{1} (1 - x^2) \, dx \). This calculates the area between those bounds.

• Each integral gives the area under the top curve minus the area under the bottom curve.
• The result is the net area between those curves over the specified interval.

Factoring Polynomials

Factoring is a crucial algebraic technique used to simplify polynomial equations and is especially important when identifying points of intersection. For instance, to find where the curves \( y = x^2 \) and \( y = 2 - x \) intersect, we solved the form \( x^2 + x - 2 = 0 \).
This quadratic equation is factored into \( (x - 1)(x + 2) = 0 \). Factoring reveals the solutions \( x = 1 \) and \( x = -2 \), highlighting where these curves cross.
Other intersections can often be found using similar techniques of setting equations equal and solving the resultant polynomial. Factoring reduces the complexity of this process, helping succinctly identify crucial solution points needed for further area calculations.