Problem 8 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=4 x^{2}-6 x, y=0$$

Step-by-Step Solution

Answer

The area between the curve and the x-axis is \( \frac{9}{4} \).

Step 1: Identify Points of Intersection

To find the area between the curve and the x-axis, identify where the parabola intersects the x-axis. Set the equation of the parabola equal to zero:\[ 4x^2 - 6x = 0 \]Factor the equation to find the values of \(x\):\[ x(4x - 6) = 0 \]This gives the solutions \(x = 0\) and \(x = \frac{3}{2}\). These points are our limits of integration.

Step 2: Set Up the Integral

With the limits of integration identified, we set up the integral to find the area. As the area is bounded by \(y = 4x^2 - 6x\) and \(y = 0\), our integral becomes:\[ \int_{0}^{\frac{3}{2}} (4x^2 - 6x) \, dx \]

Step 3: Integrate the Function

Integrate the function \(4x^2 - 6x\) with respect to \(x\):\[ \int (4x^2 - 6x) \, dx = \left( \frac{4}{3}x^3 - 3x^2 \right) + C \]Here, \(C\) is the constant of integration, which will cancel out when evaluating the definite integral.

Step 4: Evaluate the Definite Integral

Apply the limits of integration to evaluate the definite integral:\[ \left[ \frac{4}{3}x^3 - 3x^2 \right]_0^{\frac{3}{2}} = \left( \frac{4}{3}\left(\frac{3}{2}\right)^3 - 3\left(\frac{3}{2}\right)^2 \right) - \left( \frac{4}{3}(0)^3 - 3(0)^2 \right) \]Simplify the expression:- Evaluate the terms: - \( \frac{4}{3}\left( \frac{27}{8} \right) = \frac{9}{2} \) - \( 3\left( \frac{9}{4} \right) = \frac{27}{4} \)- Combine and simplify:\[ \frac{9}{2} - \frac{27}{4} = \frac{18}{4} - \frac{27}{4} = -\frac{9}{4} \]

Step 5: Take Absolute Value of the Area

Since area cannot be negative, take the absolute value:\[ | -\frac{9}{4} | = \frac{9}{4} \]

Key Concepts

Definite Integral

A definite integral is a way to find the area under a curve between two specific points called the limits of integration. Imagine a curve representing some function on a graph. The definite integral calculates the total area between that curve and the x-axis, from one point (the lower limit) to another point (the upper limit). In the exercise, this is demonstrated by setting up and solving the integral of the function from

• the lower limit, 0,
• to the upper limit, \(\frac{3}{2}\).

To perform the integration, you take the antiderivative of the function, evaluate it at the upper limit, and then subtract the value at the lower limit. This process gives you the net signed area between the curve and the x-axis. Once you have the result of the integral, if it's negative, simply take its absolute value, since the area cannot be negative.

Area Under a Curve

The phrase "area under a curve" specifically refers to the space between a curve plotting a function and the x-axis, within specified boundaries on the x-axis. This is important in calculus because it gives tangible meaning to the accumulated function over a certain interval. For functions that dip below the axis, this area is conventionally considered negative, which is why we often take absolute values to determine the physical area.In this case, the area under the parabola represented by the equation \(y = 4x^2 - 6x\) and above the x-axis is found using the integral. Since this curve is a parabola that dips below the axis between \(x = 0\) and \(x = \frac{3}{2}\), the initial calculation resulted in a negative area, \(-\frac{9}{4}\). By taking the absolute value, we confirm the non-negative size of the region, giving an actual area of \(\frac{9}{4}\).

Parabola

A parabola is a specific type of curve that has a distinct U-shape, which can also appear upside down depending on the orientation. It is characteristically described by a quadratic function, i.e., an equation where the highest degree of x is squared. For the exercise you worked on, the parabola was defined by the equation \(y = 4x^2 - 6x\).The points at which the parabola intersects the x-axis are the roots or solutions of the quadratic equation when set equal to zero. This is done through factoring, which indicated that the parabola crossed the x-axis at points \(x = 0\) and \(x = \frac{3}{2}\).Understanding a parabola's shape and intersections is crucial for determining the bounded area in integral calculus, as it lays out the exact range of x-values where we're interested in calculating the area under or above the curve.