SOLUTION: What is the approximate area of the region bounded by the graphs: y equals one-half times x squared plus 2., x = 0 , x = 3 and y = 0 (the yellow region)? Show all work to receiv

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Question 111411: What is the approximate area of the region bounded by the graphs:
y equals one-half times x squared plus 2., x = 0 , x = 3 and y = 0 (the yellow region)?
Show all work to receive credit.
The graph of y equals one-half times x squared plus 2 is drawn. The area under the graph and bounded by the lines y equals 0, x equals zero and x equals 3 is shaded yellow.
I know this sounds confusing, I couldn't figure out how to draw it. On the graph
it looks like a parabala the yellow region is on the positive side of the x axis beneath the parabala. It is divided into 3 sections. (1,0)(2,0)and(3,0)

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the approximate area of the region bounded by y+=+%281%2F2%29x%5E2%2B2, x+=+0, x+=+3, and y+=+0
First, it would be helpful to see the graph of the equation:
graph%28600%2C400%2C-5%2C5%2C-5%2C8%2C%281%2F2%29x%5E2%2B2%29
Now divide the region into two areas by a vertical line through x = 2 and another vertical line through x = 3.
You can approximate the shape of these two area by trapezoids (lying on their sides).
The first trapezoid b%5B1%5D+=+2andb%5B2%5D+=+4
How did I get these?
Just substitute x = 0 into the equation and solve for y.
y%5B1%5D+=+%281%2F2%29%280%29%5E2%2B2
y%5B1%5D+=+2 and...
y%5B2%5D+=+%281%2F2%29%282%29%5E2%2B2
y%5B2%5D+=+2%2B2
y%5B2%5D+=+4
Now you can find the area of the first trapezoid using: A+=+h%28b%5B1%5D%2Bb%5B2%5D%29%2F2
The h here is just x = 2, so...
A%5B1%5D+=+2%282%2B4%29%2F2
A%5B1%5D+=+6
For the second trapezoid, you have:b%5B1%5D+=+4 and b%5B2%5D+=+6.5 from:
y%5B1%5D+=+%281%2F2%29%282%29%5E2%2B2
y%5B1%5D+=+4 and...
y%5B2%5D+=+%281%2F2%29%283%29%5E2%2B2
y%5B2%5D+=+%281%2F2%29%289%29%2B2
y%5B2%5D+=+6.5
So, the area of the second trapezoid is:
A%5B2%5D+=+%281%29%284%2B6.5%29%2F2
A%5B2%5D+=+10.5%2F2
A%5B2%5D+=+5.25
Finally, add these two areas together to get:
A%5Bt%5D+=+6%2B5.25
A%5Bt%5D+=+11.25 as the approximate area of the enclosed region.