find the area between x-axis and the func y = -x+x^3?
358253
-x+x^3 = 0\nx^3 - x = 0\nx(x^2-1) = 0\nx(x-1)(x+1) = 0\nx = 0, x = 1, x = -1\nThis function is odd because\n-(-x) + (-x)^3 = x - x^3 = -(-x+x^3)\nsince it is odd, the area between -1 and 0 is the same as the area between 0 and 1.  You can just find one of the areas and then multiply it by two.\nIntegrate from -1 to 0 of \n-x + x^3 dx\n= -(1/2)x^2 + (1/4)x^4 from -1 to 0\n-(1/2)(0)^2 + (1/4)(0)^4 - [-(1/2)(-1)^2 + (1/4)(-1)^4]\n= 0 + 0 - (-1/2) - (1/4)\n= 1/2 - 1/4\n= 1/4\nNow multiple this by 2\n2(1/4) = 1/2
408554
yes. take out an x \nyou answer should be x(x+3)
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!
680840
Trapezoidal Rule : area occupied by given curve,y=f(x),\nbetween limits (a,b),divided into various values,x0,x1,x2,...xn,for which corresponding values are y0,y1,y2,y3....yn=f(xn).\n\nHence area =A=integration(y)=integration(f(x))\n\nA=Integration between a and b(f(x))\n\nA=(h/2)[(y0+yn)+2(y1+y2+....yn-1)]\nwhere h=(b-a)/n
342990
Take the derivative of each equation. Subtract one from the other. That'll get you the area enclosed between the two.
209253
Use this formula(it uses the third-order determinant)\n\nIf the 3 points are\n(x1,y1)\n(x2,y2)\n(x3,y3)\nThen the area A is:\n\nA = \n|x1 y1 1|\n|x2 y2 1|\n|x3 y3 1|\n\n\n\n^_^
920959
r(x)= x^3+8/x^2+4\n\nAssuming you mean r(x)= (x^3 + 8)/(x^2 + 4)\n= (x + 2)(x^2 - 2x + 4)/(x^2 + 4)\n\nWhen x = 0, r(x) = 8/4 = 2.\nSo it cuts the y- axis at (0, 2)\n\nr(x) = 0 when (x + 2)(x^2 - 2x + 4)/(x^2 + 4) = 0, ie when x = -2\n\nSo it cuts the x-axis at (-2, 0)\n\nie x-intercept is -2 and y-intercept is 2.
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
905083
plot (using your calculator)\n\ny = x^1/3 + x^1/9 - 60\n\nthe values of x where y = zero are your answers.
824521
this just means Y is a function of x\n\ny = x^2(2-3)\ny = -x^2
878479
To find the inverse function switch the x's and y's\n\ny=x^2-3\nbecomes\nx=y^2-3\nand after simplifying\ny=+/-sqrt(x+3) -- don't forget the plus or minus, you need both for this one!
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2
660602
d/dx[y/x^2] gives y'/x^2 - 2y/x^3 on expansion
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
961234
Let x = y^(1/3)\nx^2 - 9x + 8 = 0\n(x - 8)(x - 1) = 0\nx = 8 or 1\ny^(1/3) = 8 or 1\ny = 2 or 1
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
1117258
Area of a square = L * L\n\nArea of A = x * x ..........(1)\nArea of B = y *y  ..........(2)\n  \n y = x -1               ..........(3)\n\nA = 4 * B              ...........(4)\n\nSubstitute (1) & (2) into (4) gives:\n\nx^2 = 4 y^2           ...........(5)\n\nSub. (3) into (5)\n\nx^2 = 4 (x-1) ^2\n\nSolve:\n\nx^2 = 4( x^2 -2x +1)\nx^2 = 4x^2 -8x +4\nOr\n\n3x^2 - 8x +4 =0\n\nSecond order eq.\n\nx1,x2= [+8 +- Sqr(8^2 - 4*3*4)]/(2*3)\n\nx1,x2 = [8+-4]/6\n\nx1 = 2   &  x2 = 4/6\n\nThen\ny1 = 2-1 =1   & y2 = 4/6 -1 = -2/6 ( illogical )\n\nthen\n\nx= 2  and y = 1