Question

use the midpoint rule with (n = 4) subintervals to approximate the area (a) of the region bounded by the given curves. (y=sqrt3{16 - x^{3}},y = x,x = 0). (round your answer to three decimal places.)

Explanation:

Step1: Find the intersection points

First, find the intersection of $y = \sqrt[3]{16 - x^{3}}$ and $y=x$. So, $\sqrt[3]{16 - x^{3}}=x$. Cubing both sides gives $16 - x^{3}=x^{3}$, then $2x^{3}=16$, and $x = 2$. Also, we have $x = 0$ as one of the boundaries. The interval of integration is $[0,2]$.

Step2: Calculate $\Delta x$

The formula for $\Delta x=\frac{b - a}{n}$, where $a = 0$, $b = 2$ and $n = 4$. So, $\Delta x=\frac{2-0}{4}=0.5$.

Step3: Determine the sub - intervals and mid - points

The sub - intervals are $[0,0.5]$, $[0.5,1]$, $[1,1.5]$, $[1.5,2]$. The mid - points are $x_1=0.25$, $x_2 = 0.75$, $x_3=1.25$, $x_4 = 1.75$.

Step4: Evaluate the function at mid - points

The area $A$ using the Mid - point Rule is $A\approx\Delta x\sum_{i = 1}^{4}f(x_i)$. Here, $f(x)=\sqrt[3]{16 - x^{3}}-x$.
$f(0.25)=\sqrt[3]{16-(0.25)^{3}}-0.25=\sqrt[3]{16 - 0.015625}-0.25=\sqrt[3]{15.984375}-0.25\approx2.519 - 0.25=2.269$.
$f(0.75)=\sqrt[3]{16-(0.75)^{3}}-0.75=\sqrt[3]{16 - 0.421875}-0.75=\sqrt[3]{15.578125}-0.75\approx2.505-0.75 = 1.755$.
$f(1.25)=\sqrt[3]{16-(1.25)^{3}}-1.25=\sqrt[3]{16 - 1.953125}-1.25=\sqrt[3]{14.046875}-1.25\approx2.41-1.25 = 1.16$.
$f(1.75)=\sqrt[3]{16-(1.75)^{3}}-1.75=\sqrt[3]{16 - 5.359375}-1.75=\sqrt[3]{10.640625}-1.75\approx2.19-1.75 = 0.44$.

Step5: Calculate the approximate area

$A\approx\Delta x(f(0.25)+f(0.75)+f(1.25)+f(1.75))$.
$A\approx0.5(2.269 + 1.755+1.16+0.44)=0.5\times5.624 = 2.812$.

Answer:

$2.812$