Mathematics, 17.05.2021 01:00 kamand10
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Use the mid-point rule with n = 2 to approximate the area of the region bounded by y equals the cube root of the quantity 16 minus x cubed y = x, and x = 0.
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Answers: 3
![\displaystyle y=\sqrt[3]{16-x^3}, y=x\text{ and } x=0](/tpl/images/1327/2634/c7d60.png)
![\displaystyle x=\sqrt[3]{16-x^3}\Rightarrow x=2\Rightarrow (2, 2)](/tpl/images/1327/2634/197bb.png)
![\displaystyle \text{Area}=\int_{0}^2\sqrt[3]{16-x^3}-x\, dx](/tpl/images/1327/2634/76b5a.png)
![f(x)=\sqrt[3]{16-x^3}-x](/tpl/images/1327/2634/5b540.png)
![\displaystyle \text{Area}\approx(\sqrt[3]{16-(.5)^3}-(.5))+(\sqrt[3]{16-(1.5)^3}-(1.5))}\approx 2.8418](/tpl/images/1327/2634/3ba20.png)
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Use the mid-point rule with n = 2 to approximate the area of the region bounded by y equals...
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