Question

Consider the function f(x)=(x-3)^2+2. Evaluate a sum to approximate the area under the curve for the domain 0 £ x £ 2 using the type of rectangles in each part.
a. Use inscribed rectangles 0.5 units wide.
b. Use circumscribed rectangles 0.5 units wide.

Answer 1

x= 0 to 2?

Answer 2

how did the \(\leq\) turn in to a £ ??

Answer 3

not sure this question has me so confused

Answer 4

intervals are of length \(.5\) so you have the points \(0,.5,1,1.5,2\) so contend with

Answer 5

your job is to multiply base times height of 4 rectangles and add them up
the base of each of these 4 rectangles is \(.5\) and the height is what you get when you replace \(x\) by \(.5,1,1.5,2\) respectively

Answer 6

it is not that simple Sat. Since they want different sample points.

Answer 7

Use inscribed rectangles 0.5 units wide.
since your function is decreasing on \([0,2]\) use the right hand endpoints for inscribed, and the left hand endpoints for circumscribed

Answer 8

I always hated Riemann sums

Answer 9

for inscribed it is
\[f(.5)\times .5+f(1)\times .5+f(1.5)\times .5+f(2)\times .5\] or more simply put
\[\left(f(.5)+f(1)+f(1.5)+f(2)\right)\times .5\]

Answer 10

circumscribed is
\[\left(f(0)+f(.5)+f(1)+f(1.5)\right)\times .5\]

Answer 11

hope this is clear, your function \(f(x)=(x-3)^2+2\) has vertex at \((3,2)\) on your interval it is deceasing, making the left hand endpoints the largest, giving you circumscribed rectangles, and the right hand endpoints as the smallest giving you inscribed

Answer 12

so a. would be (f(.5)+f(1)+f(1.5)+f(2))×.5 and b. would be (f(0)+f(.5)+f(1)+f(1.5))×.5?

Answer 13

yes exactly, for the reason above
on \([0,2]\) function is decreasing
http://www.wolframalpha.com/input/?i=y%3D%28x-3%29^2%2B2

Answer 14

thank you! you really helped me understand the reason behind the answer and went in-depth with explaining :)

Answer 15

clear right? left hand endpoints are the largest rectangles,
each has base \(.5\)

yw