y = −x2 + 10x + 24 for −1 ≤ x ≤ 3.

How many subintervals do you need to estimate the area to within 0.1 unit^{2}?

The only clue i have is something to do with w × (h_{u} - h_{l}) ≤ 0.1

where w = width of subinterval, h_{u }= height of largest upper rectangle and h_{l }= height of smallest lower rectangle

Any ideas?

Thanks!

Guest Dec 10, 2017

#1**+2 **

y = −x2 + 10x + 24 for −1 ≤ x ≤ 3.

concave down parabola

Area under the curve.

\(y=-x^2+10x+24\\ y=-(x^2-10x-24)\\ y=-(x-12)(x+2)\\\)

zeros at x= -2 and x= 12

So there are no roots between x=-1 and x=3 and the curve is above the x axis for this entire region.

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the above stuff is only relevant if you want the absolute value of the areas, if you do not care if negative areas cancel postiive areas then you do not need to worry about it.

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Since this is a parabola, if you use simpson's Rule you will need only **one** subunit because Simpson's Rule gives exact results with parabolas. (You didn't say what numerical method you had to use)

check:

x | -1 | 1 | 3 |

y | -1-10+24=13 | -1+10+24=33 | -9+30+24=45 |

\(area=\frac{3--1}{6}(13+4*33+45)\\ area=\frac{4}{6}(190)\\ area=126.\dot6\;\;units^2\)

check:

\(area=\displaystyle\int_{-1}^3\;-x^2+10x+24\;\;dx\\ area=\left[\;\frac{-x^3}{3}+5x^2+24x\;\right]_{-1}^{3}\\ area=\left[-9+45+72\;\right] - \left[\;\frac{1}{3}+5-24\;\right]\\ area=108 - \left[-19+\frac{1}{3}\;\right]\\ area=108 +19-\frac{1}{3}\\ area=126.\dot6\;units^2\\\)

Melody
Dec 11, 2017