Problem 26 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=x^{2}+2 x-8, y=x+4$$

Step-by-Step Solution

Answer

The area bounded by the curves is 57.5 square units.

Step 1: Set up the equations

We have two curves given by the equations: \(y = x^2 + 2x - 8\) and \(y = x + 4\). We are tasked with finding the area between these curves.

Step 2: Find the points of intersection

To find the points of intersection, set the two equations equal to each other: \(x^2 + 2x - 8 = x + 4\). Simplify to get: \(x^2 + x - 12 = 0\). Solve this quadratic equation using the quadratic formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = -12\). The discriminant \(b^2 - 4ac = 1^2 - 4 \times 1 \times (-12) = 49\). So, the solutions are \(x = \frac{{-1 \pm 7}}{2}\), giving \(x = 3\) and \(x = -4\).

Step 3: Set up the integral for area

The area between the curves is given by the integral of the difference of the functions from the left intersection point to the right one. The integral is \(\int_{-4}^{3} ((x + 4) - (x^2 + 2x - 8)) \, dx\). Simplify the integrand: \(x + 4 - (x^2 + 2x - 8) = -x^2 - x + 12\).

Step 4: Evaluate the integral

Integrate \(-x^2 - x + 12\) with respect to \(x\): \(\int -x^2 \, dx = -\frac{x^3}{3}\), \(\int -x \, dx = -\frac{x^2}{2}\), and \(\int 12 \, dx = 12x\). So, the integral becomes: \(-\frac{x^3}{3} - \frac{x^2}{2} + 12x\). Evaluate this from \(-4\) to \(3\).

Step 5: Find the definite integral value

Evaluate \((-\frac{3^3}{3} - \frac{3^2}{2} + 12(3)) - (-\frac{(-4)^3}{3} - \frac{(-4)^2}{2} + 12(-4))\). Compute: \((-9 - 4.5 + 36)\) is \(22.5\), and \((\frac{64}{3} - 8 + (-48))\) is \(-\frac{104}{3}\), so the total area is \(22.5 - (-\frac{104}{3}) = \frac{67.5 + 104}{3} = 57.5\).

Step 6: Conclusion

The total area bounded by the curves \(y = x^2 + 2x - 8\) and \(y = x + 4\) is \(57.5\).

Key Concepts

Area between curves

When working with calculus, you may often need to find the area between two curves. This is incredibly useful for visualizing and understanding how two functions interact on a coordinate plane. To determine this area, you first need to understand which function lies above the other between certain points, as you will find the area of the "gap" between them.

Here’s a simple method to find the area between two curves:

• Identify the two functions: Let's say we have functions: \(f(x)\) and \(g(x)\).
• Determine the interval over which you will calculate the area. This requires finding where the curves intersect.
• Set up an integral to calculate the area of one curve minus the other, across this interval: \(\int_a^b [f(x) - g(x)] \, dx\).

This integral gives us the total area where \(f(x)\) is the upper curve and \(g(x)\) is the lower curve in the given interval. Don't forget to take the absolute value if you encounter negative areas within your calculations!

Definite integral

The definite integral is a core concept in calculus that allows you to calculate the accumulated quantity, such as area under curves, over a specified interval. It is fundamentally written as \(\int_a^b{f(x) \, dx}\), representing the accumulated "area" under \(f(x)\) from \(x = a\) to \(x = b\).

The definite integral is extensively used for:

• Finding areas under or between curves.
• Calculating the work done by a force over a distance.
• Determining the total accumulated value of changing quantities over time or space.

In practice, to solve a definite integral, you perform two main steps:

• Integrate the function \(f(x)\) to find its antiderivative \(F(x)\).
• Evaluate \(F(x)\) at the upper and lower limits of the interval \(a\) and \(b\), and find the difference \(F(b) - F(a)\).

This difference represents the net area under the curve between the specified points, simplifying the calculation of areas and sums in several practical situations.

Quadratic equations

A quadratic equation is a key mathematical tool that comes in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). Solving quadratic equations is crucial in numerous calculus problems, like finding intersection points between parabolic curves.

You can solve quadratic equations using various methods, such as:

• Factoring: When the equation can be expressed as \((x + p)(x + q) = 0\), find roots \(p\) and \(q\).
• Completing the square: Rewriting the equation in the form \((x + d)^2 = e\) to find solutions in simpler expressions.
• The quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), which directly provides the solutions for any quadratic equation, and is especially useful when other methods are not feasible.

The roots obtained using these methods tell us where the intersecting points of quadratic curves lie on the graph, which is paramount when determining the area between such curves.

Intersection points

Finding the intersection points of curves is an essential step in solving many calculus problems, especially when calculating areas bounded by curves. Intersection points are locations at which two curves intersect each other, representing shared coordinates on a graph.

To find intersection points between two equations, follow these steps:

• Set the two equations equal to each other.
• Solve for \(x\), often resulting in solving a quadratic equation.
• Substitute back the \(x\) values found into either original equation to find the corresponding \(y\) values.

These steps help you find precisely where the curves meet on the graph. Once you identify these points, you can use them to set the limits of integration when calculating the area that curves enclose with the definite integrals.