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Chapter 4: Problem 32

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

Short Answer

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

Step by step solution

01

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 \(\int_{0}^{8} (1 + \sqrt[3]{x}) dx\).
02

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 \(F(8) - F(0) = (8 + \frac{3}{4}8^{4/3}) - (0 + \frac{3}{4}0^{4/3})\).
03

Final Calculation

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

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