Question

Find the area bounded by the parabola: \(\mathrm{y}=\mathrm{x}^{2}-3 \mathrm{x}\) and the line: \(\mathrm{y}=\mathrm{x}\).

Short answer

The area bounded by the parabola \(y = x^2 - 3x\) and the line \(y = x\) is \(\frac{32}{3}\) square units.

Step-by-step solution

Find the points of intersection

To find the points of intersection between the parabola and the line, we need to set their equations equal to each other and solve for \(x\): \(x^2 - 3x = x\) First, let's move all terms to one side of the equation: \(x^2 - 3x - x = 0\) Now simplify: \(x^2 - 4x = 0\) Factoring out an \(x\), we get: \(x(x - 4) = 0\) This equation is true when either \(x = 0\) or \(x - 4 = 0\). Solving for each gives: 1. \(x = 0\) 2. \(x = 4\) Now we need to find the corresponding \(y\) values. We can use the equation \(y = x\) for this: 1. When \(x = 0\), \(y = 0\). 2. When \(x = 4\), \(y = 4\). Thus, the points of intersection are \((0, 0)\) and \((4, 4)\).

Set up the integral for the area between curves

To find the area between the curves, we'll integrate along the \(x\)-axis from the left intersection point to the right intersection point. In general, the area between two curves \(y = f(x)\) and \(y = g(x)\) from \(x = a\) to \(x = b\) can be found using the integral: \(\int_a^b (f(x) - g(x)) dx\) In our case, \(f(x) = x^2 - 3x\), \(g(x) = x\), \(a = 0\), and \(b = 4\). So, the integral is: \(\int_0^4 ((x^2 - 3x) - x) dx\) Now simplify the integrand: \(\int_0^4 (x^2 - 4x) dx\)

Evaluate the integral and obtain the area

To evaluate the integral, we first find the antiderivative of the integrand: \[\begin{aligned} \int(x^2 - 4x) dx &= \frac{1}{3}x^3 - 2x^2 + C \end{aligned}\] Now, we'll evaluate this antiderivative at the limits of integration: \[\begin{aligned} \left[\frac{1}{3}x^3 - 2x^2\right]_0^4 &= \left(\frac{1}{3}(4)^3 - 2(4)^2\right) - \left(\frac{1}{3}(0)^3 - 2(0)^2\right) \\ &= \left(\frac{64}{3} - 32\right) \\ &= \left(\frac{64}{3} - \frac{96}{3}\right) \\ &= \frac{-32}{3} \end{aligned}\] Since area cannot be negative, we take the absolute value to obtain the area: Area = \(\left|\frac{-32}{3}\right|\) = \(\frac{32}{3}\) Hence, the area bounded by the parabola \(y = x^2 - 3x\) and the line \(y = x\) is \(\frac{32}{3}\) square units.

Key concepts

Parabola

A parabola is a type of curve that is defined by a quadratic equation. In simple terms, it's a shape that looks like an arch or a U, depending on its orientation. The general form of a parabolic equation is given by \(y = ax^2 + bx + c\). Here, the parabola is represented by \(y = x^2 - 3x\).

• In this equation, the coefficient of \(x^2\) is 1, which means the parabola opens upwards.
• The terms \(-3x\) and the implied constant term of zero influence where the parabola shifts and its specific shape along the coordinate grid.

Understanding the characteristics of the parabola is essential as we examine its interaction with other lines or curves. In particular, we're interested in how and where this parabola intersects with the line \(y = x\). This is critical to finding the area between them.

Line Intersection

To find the points where a parabola intersects with a line, we set their equations equal to each other. This is because, at the intersection points, both the parabola and the line share the same \(x\) and \(y\) coordinates.

• For our example, we equate \(x^2 - 3x\) and \(x\).
• This leads to the equation \(x^2 - 4x = 0\).
• By factoring, \(x(x - 4) = 0\), we find the solutions \(x = 0\) and \(x = 4\).

These solutions represent the \(x\)-coordinates of the intersection points. Plugging these back into the equation of the line \(y = x\), we determine the complete intersection points: \((0, 0)\) and \((4, 4)\). These are crucial as they define the limits of integration for calculating the area between the two curves.

Integral Calculus

Integral calculus is a branch of mathematics dealing with integrals. It is used for finding areas, among other things. To find the area between two curves, we need to calculate the integral of the difference between the curves over a specific interval. In our exercise, we integrate from the left intersection point \(x = 0\) to the right intersection point \(x = 4\).

• The general formula to find the area between two curves \(y = f(x)\) and \(y = g(x)\) between \(x = a\) and \(x = b\) is:
  \[ \int_a^b (f(x) - g(x)) \, dx \]
• For our specific problem, this turns into:
  \(\int_0^4 ((x^2 - 3x) - x) \, dx\), which simplifies to \(\int_0^4 (x^2 - 4x) \, dx\).

This integral represents the area we are trying to find. Calculating it gives us the precise measure of the space enclosed between the parabola and the line.

Antiderivative

The antiderivative, also known as the indefinite integral, is the reverse process of differentiation. Finding an antiderivative involves determining a function whose derivative is the given function. In the context of our problem, we find the antiderivative of \(x^2 - 4x\).

• The antiderivative of \(x^2\) is \(\frac{1}{3}x^3\), as deriving the latter gives us \(x^2\).
• The antiderivative of \(-4x\) is \(-2x^2\), because the derivative of \(-2x^2\) provides \(-4x\).

Combining these results, the antiderivative of the integrand \(x^2 - 4x\) is \(\frac{1}{3}x^3 - 2x^2\).
We then evaluate this antiderivative from \(x = 0\) to \(x = 4\), applying the Fundamental Theorem of Calculus. This gives us:
\[ \left( \frac{1}{3} \times (4)^3 - 2 \times (4)^2 \right) - \left( \frac{1}{3} \times (0)^3 - 2 \times (0)^2 \right) = \frac{32}{3} \]This calculation yields the exact area, taking the absolute value to ensure a positive result, confirming the area as \(\frac{32}{3}\) square units.