Problem 33 - Essential Calculus

Question

Find the area of the region bounded by the graphs of the equations. $$ y=x^{3}+x, \quad x=2, \quad y=0 $$

Step-by-Step Solution

Answer

The area of the region bounded by the given graphs is 6 square units.

Step 1: Understanding the given equations

The given equations are \(y = x^{3} + x\), \(x = 2\), and \(y = 0\). The first equation is a cubic function, the second represents a vertical line at \(x = 2\), and the third represents the x-axis.

Step 2: Determining the limits of integration

The area of interest is bounded by \(x = 2\) and the intersection of the cubic function with the x-axis, which occurs when \(y = 0\). Solving the equation \(x^3 + x = 0\) for \(x\), obtains the roots 0 and -1. However, the area in question is in the first quadrant, indicating that the lower limit of the integration is 0 and the upper limit is 2.

Step 3: Setting up and calculating the integral

By taking vertical slices of the area, the variable of integration is \(x\), so \(dy = dx\). The height of each slice is \(y\), given by the equation \(y = x^3 + x\). Therefore, the infinitesimal area is \(dA = y * dx = (x^3 + x) * dx\). By integrating \(dA\) from 0 to 2, the area \(A\) is obtained. Hence, \(A = \int_{0}^{2} (x^3 + x) dx\). Evaluate this integral using the power rule and fundamental theorem of calculus.

Step 4: Final Calculation

By using the power rule, integrate each term separately to get \(\frac{x^4}{4} + \frac{x^2}{2}\), then apply the limits of integration. This gives, \(\frac{2^4}{4} + \frac{2^2}{2} - (0)\) , simplifying to \(\frac{4+2}{1} = 6\).