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.

Question

babbs12 · Answer

$$\sqrt[3]{16-x^3}$$

mathmate · Answer

First step:
find the end points of the region.
Can you do that?

babbs12 · Answer

the limits?

mathmate · Answer

yes! the limits of the integration, which correspond to the end points of the region, where 
$\Large \sqrt[3]{16-x^3}$ and $\Large y=x$ meet.

babbs12 · Answer

the point where they meet is 2,2 according to the graph

mathmate · Answer

Excellent, so you've drawn a graph of the two functions!
So now you know the limits of integration!

babbs12 · Answer

the limits are 0 and cbrt(16-x^3)

mathmate · Answer

That's the y-limits.
To do the mid point rule, you split the region into vertical strips, n=2 means two of them.
So the limits along x-axis are 0 and 2.

babbs12 · Answer

okay

mathmate · Answer

Do you know how the mid-point rule works?

babbs12 · Answer

its the mid point of the rectangle used in the reimann sum right?

babbs12 · Answer

middle/height

mathmate · Answer

Exactly.
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mathmate · Answer

Can you give the answer in terms of the diagram?

babbs12 · Answer

so the mid points would be .5 and 1.5

mathmate · Answer

You are approximating the area of the region with two rectangles, each of height y1 and y2.  Yes the mid-points are 0.5 and 1.5 (where the function is evaluated)

babbs12 · Answer

so the function is cbrt(16-x^2) - x ???

mathmate · Answer

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The sum of the area of the two rectangles is your answer.
Yes, that's the height of the rectangles, at x=0.5, and x=1.5.
Great!