Problem 23 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=x^{2}+5 x, y=3-x^{2}$$

Step-by-Step Solution

Answer

The area bounded by the curves is approximately 58.792 square units.

Step 1: Set the equations equal to find points of intersection

We find where the curves intersect by solving the equation\( x^2 + 5x = 3 - x^2 \). Simplify and arrange the equation as follows: \( 2x^2 + 5x - 3 = 0 \). This is a quadratic equation.

Step 2: Solve the quadratic equation

Use the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \) where \( a = 2, b = 5, c = -3 \). Calculate the discriminant \( b^2 - 4ac = 25 + 24 = 49 \). So the solutions for \( x \) are \( x = \frac{{-5 \pm 7}}{4} \), giving \( x = \frac{1}{2} \) and \( x = -3 \).

Step 3: Find y-values of intersection points

Substitute \( x = \frac{1}{2} \) into either original equation to find \( y = \frac{1}{4} + \frac{5}{2} = \frac{11}{4} \) and substitute \( x = -3 \) to get \( y = 9 - 15 = -6 \). Hence, the points are \( (-3, -6) \) and \( \left(\frac{1}{2}, \frac{11}{4}\right) \).

Step 4: Set up the integral for the area

The area between the curves is the integral of the top function minus the bottom function from \( x = -3 \) to \( x = \frac{1}{2} \). Set up the integral: \( \int_{-3}^{\frac{1}{2}} ((3 - x^2) - (x^2 + 5x)) \, dx \).

Step 5: Simplify the integrand

Simplify the integrand: \( 3 - x^2 - x^2 - 5x = 3 - 2x^2 - 5x \). The integral becomes \( \int_{-3}^{\frac{1}{2}} (3 - 2x^2 - 5x) \, dx \).

Step 6: Integrate the function

Find the antiderivative: \( \int (3 - 2x^2 - 5x) \, dx = 3x - \frac{2}{3}x^3 - \frac{5}{2}x^2 + C \). Evaluate this from \( -3 \) to \( \frac{1}{2} \).

Step 7: Calculate definite integral

Substitute and evaluate: \[ \left[ 3x - \frac{2}{3}x^3 - \frac{5}{2}x^2 \right]_{-3}^{\frac{1}{2}} = \left( \frac{3}{2} - \frac{1}{24} - \frac{5}{8} \right) - \left( -9 - \frac{54}{3} - \frac{45}{2} \right) \]. Calculate these values to determine the area.

Step 8: Solve expressions to get area

Compute: For \( x = \frac{1}{2} \), calculate value: \( \frac{36}{24} - \frac{1}{24} - \frac{15}{24} = \frac{20}{24} \). For \( x = -3 \), calculate value: \[ -9 + 18 - 67.5 = -58.5 \]. The area is \( \frac{20}{24} + 58.5 = 58.79166... \approx 58.792 \).

Key Concepts

Quadratic Equations

Quadratic equations are essential in mathematics. They are polynomials that can be written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In the exercise, we encountered a quadratic equation when we set the curves equal to each other to find their intersection points. The equation \( 2x^2 + 5x - 3 = 0 \) was solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

For our equation, \( a = 2 \), \( b = 5 \), and \( c = -3 \). Calculating the discriminant, \( b^2 - 4ac \), is crucial as it determines the nature of the roots. With a discriminant of 49, we found two real roots for \( x \), namely \( \frac{1}{2} \) and \( -3 \). Understanding how to solve quadratic equations is vital for finding intersection points of curves, a common problem in calculus.

Definite Integrals

Definite integrals are used to calculate the area under a curve over a specific interval. This concept is crucial in determining the area bounded by curves.
In the solution, we needed to calculate the integral of the difference between two functions from \( x = -3 \) to \( x = \frac{1}{2} \). We set up our integral as:

• \( \int_{-3}^{\frac{1}{2}} ((3 - x^2) - (x^2 + 5x)) \, dx \)

The process involves simplifying the expression to \( 3 - 2x^2 - 5x \) before finding the antiderivative. The definite integral provides a numerical value representing the net area between the two curves over the given interval. Evaluating a definite integral gives us a precise area, which is especially useful when determining the space between intersecting curves. In this case, calculating across two boundaries makes it easy to apply values and solve.

Area Between Curves

The concept of finding the area between curves is a fascinating application of definite integrals. When we have two curves, the area between them can be found by integrating the difference of the functions that represent the curves.
In our exercise, we determined the area between the quadratic functions \( y = x^2 + 5x \) and \( y = 3 - x^2 \). Once we identified the interval from the points of intersection, \( x = -3 \) to \( x = \frac{1}{2} \), we set up and simplified the integral to find the area:

• \( \int_{-3}^{\frac{1}{2}} (3 - 2x^2 - 5x) \, dx \)

By evaluating the integral, we found the area as approximately 58.792 square units. Understanding the area between curves is essential as it applies to many real-world problems, from calculating distances to estimating energy consumption based on varying rates.