Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,
where 
. Each interval has length 
.
At these sampling points, the function takes on values of
We approximate the integral with the Riemann sum:
Recall that
so that the sum reduces to
Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:
Just to check:
9514 1404 393
Answer:
$2.98
Step-by-step explanation:
The amount of interest earned is ...
I = P((1 +r)^t -1)
I = $9(1.1^3 -1) ≈ $2.98
Daniel will earn about $2.98 in interest in 3 years.
Answer: x intercept is -3
y intercept is -9
Step-by-step explanation: 6(0) = 2y=-18, divide both sides by 2, get -9 as a y intercept
6x+2(0)=-18
divide both sides by 6, and get -3 as an x intercept
The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b...
<span>the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. </span>
<span>2cosθ = 1 => θ = ±π/3 </span>
<span>A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ </span>
<span>= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ </span>
<span>= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. </span>
<span>.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] </span>
<span>= 3θ/2 +sin(2θ) - sin(θ) </span>
<span>Area = A(π/3) - A(-π/3) </span>
<span>= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) </span>
<span>= π.</span>
