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>
Answer:
5x-3p
Step-by-step explanation:
(1/3)(3x-6p) + 4x - p
Distribute the 1/3 to the variables in parenthesis.
x - 2p +4x - p
Combine the like variables.
5x - 3p
Hope this helps!
In this problem we need to find the value of a and b. So given that t<span>he function should be in the form f(n) = an + b and we know each value of n, then out goal is to find a and b.
For getting this purpose, we need to find a system of two equations (given that we have two unknown variables)
Therefore:
(1) f(0) = a(1) + b = 18
</span>∴ a + b = 18
<span>
(2) f(1) = a(2) + b = 24
</span>∴ 2a + b = 24<span>
Solving for a and b we have:
a = 6
b = 12
Finally:
f(n) = 6n + 12</span>
Step-by-step explanation:
Taking the first coordinate point (3,16.5)
where x= 3 and y= 16.5
optionB
Answer:
12k^5 +24k⁴+30k³
Step-by-step explanation:
the area = 6k³(2k²+4k+5)
= 12k^5 +24k⁴+30k³
