Question

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).

Answer 1

I've already calculated R3, R6, L3, M3, & M6 but I keep getting L6 wrong.

Answer 2

whats the interval ?

Answer 3

Should be 0.5 right?

Answer 4

Cause it says to use 6 rectangles

Answer 5

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

Answer 6

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 :/

Answer 7

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bi%3D0%7D%5E5+%287%2B%28-1%2Bi%2F2%29%5E2%29*1%2F2

Answer 8

O.o haha i don't get it

Answer 9

width of each interval = (2--1)/6 = 3/6 = 1/2

this is correct in your work

Answer 10

Ok :) next? What part of my calculations is wrong?

Answer 11

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

Answer 12

Right

Answer 13

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 ?

Answer 14

Yep thats right!

Answer 15

Ohhhh >.<

Answer 16

I said 7, didn't I? -.-

Answer 17

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

Answer 18

Right!!

Answer 19

so area = 1/2*[f(-1) + f(-1/2) + f(0) + f(1/2) + f(1) + f(3/2)]

Answer 20

Exactly! Thats true! Such a silly mistake :D lol

Answer 21

But I'm still getting it wrong :/

Answer 22

try entering 23.375

Answer 23

or 187/8

Answer 24

Thats right! :D

Answer 25

Thanks so much!!! <3

Answer 26

yw!

Answer 27

:)