Problem 2 - Calculus Volume 1

Question

For the following exercises, determine the area of the region between the two curves in the given figure by integrating over the \(x\) -axis. \(y=x^{2}\) and \(y=3 x+4\).

Step-by-Step Solution

Answer

The area is \(\frac{33}{2}\).

Step 1: Find Points of Intersection

To find the area between the curves, we need to determine where the curves intersect. We set the equations equal to find these intersection points:\[ x^2 = 3x + 4 \]Rearrange this equation to:\[ x^2 - 3x - 4 = 0 \]Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a = 1\), \(b = -3\), and \(c = -4\): \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \]\[ x = \frac{3 \pm 5}{2} \]This gives solutions \(x = 4\) and \(x = -1\). Thus, the curves intersect at \(x = -1\) and \(x = 4\).

Step 2: Set Up the Integral for Area

The area between the curves from \(x = -1\) to \(x = 4\) is found using the integral of the upper curve minus the lower curve. Here, the line \(y = 3x + 4\) is above the parabola \(y = x^2\) in this interval. The area \(A\) is given by:\[ A = \int_{-1}^{4} [(3x + 4) - (x^2)] \, dx \]

Step 3: Simplify the Integral Expression

Simplify the expression inside the integral:\[(3x + 4) - (x^2) = -x^2 + 3x + 4\]Now the integral becomes:\[ A = \int_{-1}^{4} (-x^2 + 3x + 4) \, dx \]

Step 4: Integrate the Expression

Perform the integration:\[ A = \left[ -\frac{x^3}{3} + \frac{3x^2}{2} + 4x \right]_{-1}^{4} \]

Step 5: Evaluate the Definite Integral

Evaluate the definite integral:Plug in the upper limit (\(x=4\)):\[-\frac{4^3}{3} + \frac{3(4)^2}{2} + 4(4) = -\frac{64}{3} + 24 + 16 = \frac{40}{3}\]Plug in the lower limit (\(x=-1\)):\[-\frac{(-1)^3}{3} + \frac{3(-1)^2}{2} + 4(-1) = \frac{1}{3} + \frac{3}{2} - 4 = -\frac{19}{6}\]Subtract the lower limit from the upper:\[\frac{40}{3} - \left(-\frac{19}{6}\right) = \frac{40}{3} + \frac{19}{6} = \frac{80}{6} + \frac{19}{6} = \frac{99}{6} = \frac{33}{2}\]The area is \(\frac{33}{2}\).

Step 6: Final Answer

The area between the curves \(y=x^2\) and \(y=3x+4\) from \(x = -1\) to \(x = 4\) is \(\frac{33}{2}\).

Key Concepts

Integration

Integration is a fundamental concept in calculus. It allows us to find areas under curves, among other applications. In problems involving the area between two curves, integration helps calculate the net area confined by the curves.

In general, when you want to find the area between two functions, you integrate the difference of these functions over a specific interval. This means subtracting the value of the lower function from the upper function, across the interval where they intersect. For example, if you have functions \(f(x)\) and \(g(x)\), the area \(A\) between them from \(a\) to \(b\) is:

• \(A = \int_{a}^{b} [f(x) - g(x)] \, dx\)

Always make sure to determine which curve lies above and which one is below in the region of interest, as this affects how you set up your integral.

• In our exercise, \(y = 3x + 4\) is the upper function, and \(y = x^2\) is the lower function.

Another important aspect of integration is its rules for finding antiderivatives. Functions power \(n\), for example, integrate as follows:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

In definite integrals, however, the constant \(C\) vanishes, as you are calculating a specific area.

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\). They are called quadratic because they involve the square of the variable \(x\). Solving these equations is crucial to many mathematical applications, including finding where two curves intersect in geometry.

In this context, the solution of the quadratic equation tells us the points \(x\) where two curves meet. For example, to find the points of intersection for \(y = x^2\) and \(y = 3x + 4\), we set the equations equal to each other:

• \(x^2 = 3x + 4\)

Rearranging gives us the standard quadratic equation:

• \(x^2 - 3x - 4 = 0\)
• Here, \(a = 1\), \(b = -3\), and \(c = -4\).

To solve it, you can use the quadratic formula:

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

The solutions give us the \(x\) values where the two functions intersect. It helps to graph these functions for a visual understanding of where they meet.

Definite Integral

The definite integral is a crucial concept in calculus used to calculate the exact area under a curve or between curves. It is represented by an integral sign with specified upper and lower limits.

• In our exercise, we use the definite integral from \(x = -1\) to \(x = 4\) to find the area between two curves.

When you set up a definite integral, you begin by identifying the curves involved and the bounds of integration.

The definite integral of a function \(f(x)\) from \(a\) to \(b\) is given by:

• \(\int_{a}^{b} f(x) \, dx\)

Unlike indefinite integrals, definite integrals compute a specific numeric value representing an area, hence no constant of integration \(C\) is added.

For our problem, the integral \(\int_{-1}^{4} (-x^2 + 3x + 4) \, dx\) finds the net area between the curves. When you evaluate this integral, you use the fundamental theorem of calculus, which involves finding the antiderivative and evaluating it at the bounds:

• \[ \left[ F(x) \right]_{a}^{b} = F(b) - F(a) \]
• Substitute and compute at the limits \(x=4\) and \(x=-1\).

This gives the precise measure of the space between the curves across the interval.