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Before performing any calculation it's good to recall a few properties of integrals:
So we apply the first property in the first expression given by the question:
![\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}](https://tex.z-dn.net/?f=%5Csmall%20%5Csf%7B%5Clongrightarrow%5Cint%20%5E3_%7B-2%7D%20%5B2f%28x%29%20%2B2%5Ddx%3D%202%20%5Cint%20%5E3%20_%7B-2%7D%20f%28x%29%20dx%2B%20%5Cint%20f%5E3%20_%7B2%7D%202dx%3D18%7D)
And we solve the second integral:
Then we take the last equation and we subtract 10 from both sides:
And we divide both sides by 2:
Then we apply the second property to this integral:
Then we use the other equality in the question and we get:
We substract 8 from both sides:
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(x - 2)²(x+1)
(x-2)(x-2)
multiply the two brackets together
(x)(x)=x^2
(x)(-2)=-2x
(-2)(x)=-2x
(-2)(-2)=4
x^2-2x-2x+4
x^2-4x+4
(x^2-4x+4)(x+1)
multiply the brackets together
(x^2)(x)=x^3
(x^2)(1)=x^2
(-4x)(x)=-4x^2
(-4x)(1)=-4x
(4)(1)=4
x^3+x^2-4x^2-4x+4x+4
Answer:
x^3-3x^2+4
<span>x+y=30.....(1) and x-y=2......(2) so(1)+(2)......x+y+x-y=30+2 so 2x=32...x=16 now apply x=16 in (1) so we get......16+y=30 so .......y=14</span>
(0, 0) → t = 0, w = 0 → w = 0 · 0
(1, 4) → t = 1, w = 4 → w = 1 · 4
(2, 8) → t = 2, w = 8 → w = 2 · 4
(3, 12) → t = 3, w = 12 → w = 3 · 4
<h3>Answer: w = 4t</h3>
shorter side : longer side = 3 : 7
Equate 3 parts tp 9 inches:
3 parts = 9 inches
Find 1 part:
1 part = 9 ÷ 3 = 3 inches
Find 7 parts:
7 parts = 3 x 7 = 21 inches
Area = Length x Width = 21 x 9 = 189 in²
Answer: The longer side is 21in and the area is 189 in²
