Split up the integration interval into 4 subintervals:
![\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac%5Cpi8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi8%2C%5Cdfrac%5Cpi4%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi4%2C%5Cdfrac%7B3%5Cpi%7D8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%7B3%5Cpi%7D8%2C%5Cdfrac%5Cpi2%5Cright%5D)
The left and right endpoints of the
-th subinterval, respectively, are


for
, and the respective midpoints are

We approximate the (signed) area under the curve over each subinterval by

so that

We approximate the area for each subinterval by

so that

We first interpolate the integrand over each subinterval by a quadratic polynomial
, where

so that

It so happens that the integral of
reduces nicely to the form you're probably more familiar with,

Then the integral is approximately

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.
Answer:
12
Step-by-step explanation:
Multiply both sides of the equation by 8/5.
4x-16=32
Move 16 to the other side
4x=32+16
add numbers on the right side
4x=48
divide both sides by 4
x=12
Answer:
when squeaky stands up to the mean girls
Step-by-step explanation:
I believe it is D)3x+2y<7
Hope I helped