Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 34

Estimate the area of the region bounded by the curves.

$y = \sqrt[3]{16 - x^3}, y = x, x = 0$ by using the Midpoint Rule with $n = 4$.

By graphing the function,







We can use a vertical strip to determine the equation of the area of the curve. So..

$\displaystyle A = \int^2_0 \left[\sqrt[3]{16 - x^3} - x\right] dx$

If we evaluate the area using Midpoint Rule, we first get the thickness of each rectangular strip. So..

$\displaystyle \Delta x = \frac{2 - 0}{4} = 0.5$

Thus, the endpoints of the five sub-intervals are $0, 0.5, 1, 1.5$ and $2$. So, the midpoints are $\displaystyle \left( \frac{0 + 0.5}{2} \right) = 0.25, \left( \frac{0.5 + 1}{2} \right) = 0.75, \left( \frac{1 + 1.5}{2} \right) = 1.25$ and $\displaystyle \left( \frac{1.5 + 2}{2} \right) = 1.75$.

Therefore, the Midpoint Rule gives


$
\begin{equation}
\begin{aligned}

\int^2_0 \left[\sqrt[3]{16 - x^3} - x\right] \approx & \Delta x [f(0.25) + f(0.75) + f(1.25) + f(1.75) ]
\\
\\
\approx & \frac{1}{2} [2.269 + 1.7475 + 1.1628 + 0.4495]
\\
\\
\approx & 2.8144 \text{ square units}

\end{aligned}
\end{equation}
$