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.
From this picture you can calculate coordinates of triangles' vertices:
.
Since
you can conclude that points A, B, C are rotated by
about the point (-0.5,0) to form points J, G, H, respectively.
Answer: A is correct choice.
Answer:
52 : 14
Step-by-step explanation:
multiply the numbers by 2
You may need to sit down with your parents or with your teacher and
go over how to add and subtract fractions.
1). "Perimeter" means the distance all the way around the square.
With a square, all 4 sides are the same length. With <u>this</u> square,
every side is 1-1/4 inches long.
Perimeter = length of all 4 sides= (1-1/4) + (1-1/4) + (1-1/4) + (1-1/4) =
(1 + 1 + 1 + 1) + (1/4 + 1/4 + 1/4 + 1/4) =
4 + 4/4 = <em>5 inches</em> .
2). (2-3/8) + (1-7/8) = (2 + 1) + (3/8 + 7/8) =
(3) + (10/8) =
3 + 1-1/4 = <em>4-1/4 .</em>
3). The difference is (1-1/6) minus (5/6) .
Before you start to do the subtraction, write the (1-1/6) as (7/6) .
Then the subtraction is (7/6) - (5/6) = 2/6 = <em>1/3</em> .
4). This one is almost the same kind of problem as #3.
It's another subtraction.
If you need (2-1/4) all together, and you already have (1-3/8),
then the amount you still have to find, or borrow, or buy, is the
difference between those two numbers.
(2-1/4) minus (1-3/8) .
The trick is to write the (2-1/4) in some form that you'll be able to
subtract (1-3/8) from it. When I learned how to do that, it was called
'borrowing', but I think now it's called 'regrouping'.
We need to work on (2-1/4):
-- take 1 from the 2, and change it into fourths.
2-1/4 = 1 and 4/4 and 1/4 = 1 and 5/4
-- Now, take those 5/4, and turn them into eighths.
Each fourth makes 2 eighths. So 5/4 = 10/8.
Now, the (2-1/4) has turned into 1-10/8 .
We did NOT change the value. It's still the same amount
as 2-1/4 , but it's just written in a different way.
And now the subtraction is easy:
(2-1/4) minus (1-3/8) =
(1-10/8) minus (1-3/8) = (zero and 7/8).
You need <em>7/8 inch</em> more string than you already have.
