Problem 12 - Calculus: Early Transcendental Functions

Question

Write the given (total) area as an integral or sum of integrals. The area above the \(x\) -axis and below \(y=4 x-x^{2}\).

Step-by-Step Solution

Answer

The total area under the curve \(y=4x-x^{2}\) from \(x = 0\) to \(x = 4\) can be expressed as the integral \(\int_{0}^{4} (4x-x^{2}) dx\).

Step 1: Identify the limits of the integral

The bounding points are the values of \(x\) where the given equation equals 0, which are \[0 = 4x-x^{2} \Rightarrow x (4 - x) = 0 \]. This yields two solutions: \(x = 0\) and \(x = 4\). These are the limits of the integral.

Step 2: Writing the integral

The area under the curve can be found by taking the integral of the curve from the lower limit \(x = 0\) to the upper limit \(x = 4\). So it is to be written as follows: \[\int_{0}^{4} (4x-x^{2}) dx\].

Step 3: Calculating the integral

Gather the constants out of the integral and then integrate each term to find its primitive function: \[\int_{0}^{4} (4x-x^{2}) dx = \left[2x^{2} - \frac{1}{3}x^{3}\right]_{0}^{4}\].

Key Concepts

Finding Limits of Integration

When dealing with definite integrals, the first step is to identify the limits of integration. These boundaries determine where on the x-axis our integration, thereby the area calculation, starts and ends.
To find these limits, especially when given a curve, set the function equal to zero. The points where it intersects the x-axis form the lower and upper limits of integration.

• For instance, with the curve described by the function \( y = 4x - x^2 \), setting \( y = 0 \) gives us the equation \( 4x - x^2 = 0 \).
• Solving \( x(4 - x) = 0 \) reveals the solutions \( x = 0 \) and \( x = 4 \), our integration limits.

These points show the sections over which we will calculate the area under the curve. They are critical as they ensure we only integrate over the specific section in question.

Area Under a Curve

Finding the area between a curve and the x-axis involves integrating the function within its limits. This process helps in various applications, such as calculating distances, volumes, or net changes.

• The integral essentially sums up the infinitely small slices of area between the curve and the x-axis over a specific interval.
• In our case, we calculate the integral from the leftmost point \( x = 0 \) to the rightmost point \( x = 4 \).

By writing the integral as \( \int_{0}^{4} (4x - x^2) \ dx \), we represent the area under the curve between these two x-values. It is crucial to interpret the function correctly since we are looking for the area above the x-axis, ensuring that all values are positive over the interval.

Polynomial Integration

Integrating polynomials involves finding the anti-derivative, which gives us a new function that represents accumulated area. This is usually straightforward, provided we follow some basic rules.

• Each term in the polynomial is handled separately: increase the exponent by one and divide by the new exponent.
• For the integral of \( 4x \), we increase the exponent to 2, giving us \( 2x^2 \).
• For \( -x^2 \), we apply the same approach to get \(-\frac{1}{3}x^3\).

After integrating, we find the area by substituting back the limits of integration. Plug these limits into the integrated function and subtract: \[\left[2x^{2} - \frac{1}{3}x^{3}\right]_{0}^{4} = \left(2(4)^2 - \frac{1}{3}(4)^3\right) - \left(2(0)^2 - \frac{1}{3}(0)^3\right) \]. This calculation gives us the precise value of the area under the curve within the specified bounds.