Split up the interval [0, 2] into 4 subintervals, so that
![[0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right]](https://tex.z-dn.net/?f=%5B0%2C2%5D%3D%5Cleft%5B0%2C%5Cdfrac12%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac12%2C1%5Cright%5D%5Ccup%5Cleft%5B1%2C%5Cdfrac32%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac32%2C2%5Cright%5D)
Each subinterval has width
. The area of the trapezoid constructed on each subinterval is
, i.e. the average of the values of
at both endpoints of the subinterval times 1/2 over each subinterval
.
So,


Answer:
im guessing its d
Step-by-step explanation:
Is it the area of the whole figure or the shaded part
Answer:
32
Step-by-step explanation:
that just is the answer use pyth theorem them add 8
Answer:
1. A possible solution for x +y is 9
Step-by-step explanation:
A possible solution for m-n is 0