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