I keep getting this wrong, can I please get some help xD Trying to estimate the area under the graph f(x) = 7 + x^2 using six approximating rectangles and left endpoints (L6).

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I keep getting this wrong, can I please get some help xD Trying to estimate the area under the graph f(x) = 7 + x^2 using six approximating rectangles and left endpoints (L6).

Mathematics
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I've already calculated R3, R6, L3, M3, & M6 but I keep getting L6 wrong.
whats the interval ?
Should be 0.5 right?

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Cause it says to use 6 rectangles
Ohhh I didn't give you the whole question! Forgive me haha! Here it is... Estimate the area under the graph of f(x)=7+x2 from x=−1 to x=2
I get this... \[L _{6}= (7/2)+(7.25/2)+(7/2) +(7.25/2)+(7/2)+(1/2)(7+1.5^2)\] But apparently I'm doing something wrong :/
http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bi%3D0%7D%5E5+%287%2B%28-1%2Bi%2F2%29%5E2%29*1%2F2
O.o haha i don't get it
width of each interval = (2--1)/6 = 3/6 = 1/2 this is correct in your work
Ok :) next? What part of my calculations is wrong?
next, you're splitting the given interval (-1, 2) into 6 subinterval sand evaluating the f(x) at the left end point in each subinterval
Right
f(x) = 7 + x^2 the first subinterval starts at -1, so the height of first rectangle is 7 + (-1)^2 = 7+1 = 8 yes ?
Yep thats right!
Ohhhh >.<
I said 7, didn't I? -.-
the second subinterval starts at x = -1 + 1/2 = -1/2 so you need to evaluate f(-1/2) for the height of second subinterval
Right!!
so area = 1/2*[f(-1) + f(-1/2) + f(0) + f(1/2) + f(1) + f(3/2)]
Exactly! Thats true! Such a silly mistake :D lol
But I'm still getting it wrong :/
try entering 23.375
or 187/8
Thats right! :D
Thanks so much!!! <3
yw!
:)

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