Problem 26 - Calculus Single Variable

Question

Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=x^{2}+5 x+9 \quad g(x)=-x^{2}+19 x-15 $$

Step-by-Step Solution

Answer

The area of the bounded region is approximately 7.67 square units.

Step 1: Find Points of Intersection

To find where the graphs intersect, set the equations equal: \(x^2 + 5x + 9 = -x^2 + 19x - 15\). Combine like terms to form the quadratic equation: \(2x^2 - 14x + 24 = 0\). Divide by 2 to simplify: \(x^2 - 7x + 12 = 0\). Factor the equation: \((x - 3)(x - 4) = 0\). The solutions are \(x = 3\) and \(x = 4\).

Step 2: Setup the Integral for Area

The bounds for integration are the points of intersection, \(x = 3\) and \(x = 4\). We integrate the difference of the functions: \(\int_{3}^{4} [( -x^2 + 19x - 15) - (x^2 + 5x + 9)]\,dx\).

Step 3: Simplify the Integrand

Simplify the expression inside the integral: \([-x^2 + 19x - 15 - x^2 - 5x - 9]\) which simplifies to \(-2x^2 + 14x - 24\).

Step 4: Compute the Definite Integral

Integrate: \(\int_{3}^{4} (-2x^2 + 14x - 24)\,dx\). Calculate each term separately: \[-\frac{2}{3}x^3 + 7x^2 - 24x\] evaluated from 3 to 4.

Step 5: Evaluate the Integral from 3 to 4

Calculate the antiderivative at the bounds:\(-\left[\frac{2}{3}(4^3) + 7(4^2) - 24(4)\right] + \left[\frac{2}{3}(3^3) + 7(3^2) - 24(3)\right]\).Simplify each term to find:\(\left[-\frac{128}{3} + 112 - 96\right] - \left[-\frac{54}{3} + 63 - 72\right].\)

Step 6: Calculate the Numerical Values

Compute separately: - For \(x = 4\): \(-\frac{128}{3} + 112 - 96 = -\frac{32}{3}\).- For \(x = 3\): \(-\frac{54}{3} + 63 - 72 = -3\).Subtract: \(-\frac{32}{3} - (-3) = -\frac{32}{3} + 3 = -\frac{23}{3}.\) However, because we're considering absolute value for area: \(\frac{23}{3}.\)

Step 7: Determine the Area

The area is the absolute value of the computed integral: \(\frac{23}{3} = 7.67.\) Square units.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that allows us to calculate the accumulated quantity, such as area under a curve, between two points on the x-axis. It is represented by the integral symbol with upper and lower bounds.
For example, in the exercise, the definite integral from 3 to 4 \[\int_{3}^{4} f(x) \, dx\] is used to find the area between the curves by calculating the net accumulation of the height of the curves between these points.
The result of a definite integral is a real number, which can also be negative, but when calculating areas, we take the absolute value to ensure a positive result.
Here, the definite integral represents the net area between the curves \(f(x)\) and \(g(x)\) within the specified bounds.

Area Between Curves

Finding the area between two curves involves integrating the difference of the two functions. This means taking the higher curve's equation and subtracting the lower curve's equation before integrating.
In the given problem, we have two functions: \(f(x) = x^2 + 5x + 9\) and \(g(x) = -x^2 + 19x - 15\). To calculate the area between these two, we find:
\[\int_{3}^{4} [(g(x)) - (f(x))] \, dx\]This calculation measures the space "trapped" between these curves from \(x = 3\) to \(x = 4\). By subtracting \(f(x)\) from \(g(x)\), we know exactly how much one curve rises above the other, which we then integrate over the specified interval.
This gives the exact net area that lies between the curves over the range we are interested in.

Points of Intersection

Points of intersection occur where the graphs of two functions meet, which means they have the same y-value for the same x-value. This is one of the key steps when finding the area between curves, as it helps define the limits for integrating.
In the exercise, to find the points of intersection, we set \(f(x) = g(x)\) and solve the resulting equation, giving us the x-values where the graphs cross each other. The solutions to the quadratic equation \((x - 3)(x - 4) = 0\) give us \(x = 3\) and \(x = 4\).
These values define the interval over which we need to integrate to find the area between the curves. Hence, they are crucial because they limit or bound the problem to the specific region where the two curves enclose an area.