Problem 32 - Essential Calculus

Question

Find the area of the region bounded by the graphs of the equations. $$ y=1+\sqrt[3]{x}, \quad x=0, \quad x=8, \quad y=0 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the given equations is 14 square units.

Step 1: Finding the Definite Integral

The area A of the region bounded by the curve \(y = f(x)\) and the x-axis, from \(x = a\) to \(x = b\) is given by the definite integral \(\int_{a}^{b} f(x) dx\). In this case, \(f(x) = 1 + \sqrt[3]{x}\), \(a = 0\), and \(b = 8\). So, to find the area A, you have to calculate \(\int_{0}^{8} (1 + \sqrt[3]{x}) dx\).

Step 2: Calculating the Integral

The integral \(\int_{0}^{8} (1 + \sqrt[3]{x}) dx\) can be computed by finding the antiderivative F of \(1 + \sqrt[3]{x}\) and then applying the fundamental theorem of calculus, which states that \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). The antiderivative of \(1 + \sqrt[3]{x}\) is \(x + \frac{3}{4}x^{4/3}\). Therefore, the area A equals \(F(8) - F(0) = (8 + \frac{3}{4}8^{4/3}) - (0 + \frac{3}{4}0^{4/3})\).

Step 3: Final Calculation

After simplifying the expression given in the previous step, you obtain that the area A equals \(8 + \frac{3}{4}8^{4/3} - 0 = 8 + 6 = 14\).