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Mathematics 21 Online
OpenStudy (anonymous):

Use inscribed rectangles to approximate the area under f(x) = –x2 + 7 for the –2 ≤ x ≤ 0 and rectangle width 0.5. 9.30 units2 9.75 units2 10.25 units2 10.75 units2

OpenStudy (anonymous):

@phi

OpenStudy (misty1212):

HI!!

OpenStudy (anonymous):

Hey

OpenStudy (misty1212):

this is going to take a lot of calculation

OpenStudy (anonymous):

ok

OpenStudy (misty1212):

first we need the endpoints of the intervals when we divide \([-2,0]\) in to pieces of length \(0.5\)

OpenStudy (anonymous):

Ok would that be -2, -1.5, -0.5 and 0?

OpenStudy (misty1212):

yes except you skipped one i think

OpenStudy (anonymous):

-1

OpenStudy (misty1212):

right

OpenStudy (anonymous):

What do I do next?

OpenStudy (misty1212):

ok next we see that it says "inscribed rectangles" it is clear that on the interval \([-2,0]\) your function \(-x^2+7\) is going up (increasing)?

OpenStudy (anonymous):

Yes?

OpenStudy (misty1212):

lol it is a parabola that opens down with vertex at \((0,7)\) so it should be clear |dw:1453211685987:dw|

OpenStudy (anonymous):

ok

OpenStudy (misty1212):

we need to know that so that we know the "inscribed rectangle" will use the left hand endpoints of your divided up intervals, not the right hand ones

OpenStudy (anonymous):

So like on the negative side?

OpenStudy (misty1212):

the interval you were given is \([-2,0]\) so yes, that is on the left of the graph, but that is not what i meant in the comment i just wrote let me try to draw a picture

OpenStudy (misty1212):

|dw:1453211840476:dw|

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