What is the area between the 4 curves: y=x^2-x-6, y=x^2-x-4, x=1, x=6?
928287
y = int from 1 to 6 [f(x) - g(x)] dx\ny= int 1,6 [x^2-x-4] dx - int 1,6 [x^2-x-6] dx\n\nint 1,6 [x^2-x-4] dx\n[(x^3)/3] - [(x^2)/2] - 4x [1,6\ninputing 6 and 1 for x you get\n[216/3] -18-24 -([1/3] - [1/2] - 4) = 34 1/6 or 205/6\n\nint 1,6 [x^2-x-6] dx\n[(x^3)/3] - [(x^2)/2] - 6x\ninputing 6 and 1 for x you get\n[216/3] -18-36 -([1/3] - [1/2] - 6) = 24 1/6 or 145/6\n\nso you get 34 1/6 - 24 1/6 = 10
704730
x(6)^2/4=117\n=>36x/4=117\n=>9x=117\n=>x=117/9=13
493520
The length of the curve between the two points according to my calculation was:\n\nd = (3/2a) + (a*ln(2)/4)\n\nThe length is minimum under the condition that\n\na = √(6/ln(2))\n\n\n\nI don't have the time to show the full calculations, but I'll give you some key points:\n\n1. Find the line element 'ds' using the formula:\n   ds = √(dx^2 + dy^2)\n   -or-\n   ds = dx * √(1 + (dy/dx)^2)\n\n2. Evaluate (dy/dx) and substitute in above equation. \n\n3. Integrate the line element 'ds' determined above from x=1 to x=2 - this should give you the length of the curve in terms of 'a'\n\n4. To find the minimum length, differentiate the equation for length with respect to 'a', make the equation equal zero & solve for 'a'. This should give you the values for 'a' that is either for minimum or maximum. Now, differentiate the new equation once more & substitute the different values of 'a'. For each case, check if the equation yields a positive value - this means that it is a minimum case.
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
766536
(x^4-y^4)\n=(x^2+y^2)(x^2-y^2)\n=(x^2+y^2)(x+y)(x-y)\n\n2.(x^4-1)\n=(x^2+1)(x^2-1)\n=(x^2+1)(x+1)(x-1)
965380
(x - 3)² + (y + 2)² = 4²\n(C): x² + y² - 6x + 4y - 3 = 0
592617
the area = (1/2)*(4+x)*(16-(1/4)*(x-4)^2)^(1/2).  you now take the derivative and set it equal to zero and solve for x.  this will give you the length of the forth side that would mazimize the area.  the derivative = (32+4x-x^2)/(2*(48+8x-x^2)^(1/2)) .  when you set this to zero and solve for x, you get that x=8....I got the derivative by applying the product rule of differentiation.  the math is a bit to long to type out in my answer
323605
Area of a circle = pi x r^2 =3.1416 x 4^2=50.265 cm^2\nArea of a circle = pi x (d^2)/4 = 3.1416 x (8^2)/4 = 50.265 cm^2\n\nCircumference of a circle = pi x diameter = 3.1416 x 8 =25.133 cm\nCircumference of a circle = pi x 2 x radius = 3.1416 x 2 x 4 =25.133 cm
868167
In polar coordinates area is\n\nA = ½∫r² dθ\n\nThe area of the sector bounded by the curve r = f(θ) and the two radial lines defined by θ1 and θ2 is\n\n½∫[θ = θ1 to θ2] f(θ)² dθ\n\nIn your problem f(θ) = 2 cos 3θ. The curve starts when θ = 0 at (2,0) and continues. It first meets (0,0) when 2 cos 3θ = 0, or θ = π/6 making a region of half a loop.\n\nSo, area of one loop =\n\n2 × ½∫[θ = 0 to π/6] ( 2 cos 3θ )² dθ\n\n= 4∫[θ = 0 to π/6] cos² 3θ dθ\n\n= 4∫[θ = 0 to π/6] ½ ( 1 + cos 6θ ) dθ\n\n= 2 { ∫[θ = 0 to π/6]  dθ + ∫[θ = 0 to π/6] cos 6θ dθ }\n\n= 2 { [θ = 0 to π/6: θ]  + 0 } because the area over the range 0 to π/6 of cos 6θ cancels to 0\n\n= π / 3
808787
2(x+4)-6=-6(x+4)-10\nadding 6\n2(x+4)=-6(x+4)-4\ndividing by 2\n(x+4)=-3(x+4)-2\nx+4=-3x-12-2\nx+4=-3x-14\nadding -4\nx=-3x-18\nadding 3x\n4x=-18\ndividing by 4\nx=-9/2
827502
1.6a^2b^2(1+3ab)\n\n2.(y-4)(y-1)\n\n3.(2b-1)(b+7)\n\n4.3(x^2-9y^2)\n=3(x+3y)(x-3y)
744242
let y=4x(x+x^1/2)^1/2\nlny=ln4+lnx+1/2ln(x+x^1/2)\ny'/y=1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)\ny'=y[1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)]
869879
Complete the square:\nx^2 - 4x + 4 + y^2 + 6y + 9 = 12+4+9\n(x-2)^2 + (y+3)^2 = 25 = 5^2\nRadius 5, centre (2,-3).
317924
See previous answer--\n\nGives vertical height as rt27\n\nTherefore 1/2*6*rt27 is area of one side\n\nBase is 6*6 = 36
993823
Between any two points there exists a line, and between three points you can find a parabola.\n\nSo, you have a parabola of the form:\nc*(x-a)*(x-b)\nSince 4,0 is a point, you can let b = 4\nso the parabola is c*(x-a)*(x-4)\nsubbing in the other points, we get:\n-3 c + 3 c a = 3\n12 c + 6 c a = 1\nSolving gives\nc = -5/18\na = -13/5\nSo the curve is:\n-5/18*(x+13/5)*(x-4)\n=-5/18x^2 + 7/18 x + 26/9
632855
4. x^2 - 3x <= 0\n\nif you factor:\n\nx(x - 3) = 0\n\nwhich means either\nx = 0 or x-3=0\nx=0 or x=3 these are the endpoints represented on the graph. if you want the area enclosed (underneath) the graph, it has to be <= 0