(a) Estimate the area under the graph of f(x) = 25 - x2 from x = 0 to x = 5 using five approximating rectangle
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Since you have to divide the area into five rectangles, you will need rectangles from:\n\n0 to 1, 1 to 2, 2 to 3, 3 to 4, 4 to 5\n\nSince you are using the right endpoints, you need to be using the second value of each division; that is, in the "x to y" examples, you'll be using the y endpoints 1,2,3,4, and 5.\n\nThe area of each rectangle will then be 1*f(x), since the width of each rectangle is 1.  Therefore, the total sum formula should be as follows (with * representing multiplication).\n\n1 * f(1) + 1*f(2) + 1*f(3) + 1*f(4) + 1*f(5)\n\nOr, quite simply, f(1) + f(2) + f(3) + f(4) + f(5) to solve using the right endpoints.  Calculating this, we get\n\n(25 - 1^2) + (25 - 2^2) + (25 - 3^2) + (25 - 4^2) + (25 - 5^2)\n(24) + (21) + (16) + (9) + 0 = 70\n\nTo calculate using the left endpoints, recall that our intervals for the rectangles were:  0 to 1, 1 to 2, 2 to 3, 3 to 4, and 4 to 5.  In this case, we use the LEFT endpoints and take the FIRST value of each interval; that is, 0,1,2,3, and 4.  The width of the rectangle doesn't change at all (the width is 1), so our left endpoint summation formula because\n\n1 * f(0) + 1 * f(1) + 1 * f(2) + 1 * f(3) + 1 * f(4)\n\nor, quite simply\n\nf(0) + f(1) + f(2) + f(3) + f(4).  We calculated f(1) through f(4) already above, so we'll copy those values.  But f(0) = (25 - 0^2) = 25, so our total now becomes:\n\n25 + 24 + 21 + 16 + 9 = 95.\n\nArea using right endpoints = 70\nArea using left endpoints = 95
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Consider the function f as mapping the points x of a set X to the points y of a set Y. Draw a horizontal line (the x axis) so that each point on the line represents an element of the set X. Draw a perpendicular line (the y axis) to intersect the x axis at the null point or zero. Each point on the y axis represents an element of the set Y. \n\nFor each x in X calculate the value of y= f(x) in Y. This gives a pair of co-ordinates (x,y). (Depending on the function, you may have more than on value of y for each x eg the function f(x)=square root of x will give + and - values.)\n\nFor any x on the x axis draw a line parallel to the y axis to intersect the x axis at x. Draw another line to intersect the y axis at the point y=f(x) and parallel to the x axis. Where these lines intersect is a point on your graph. The complete graph consists of all such points.\n\nWhere the set X is infinite (eg the set of real numbers) you cannot calculate the value of f(x) individually for every x but by calculating the values for a series of points at suitable intervals you will see a pattern emerging to the graph and you will be able to fill in the gaps by joining the (x,y) points, hopefully with a smooth and continuous line.
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Assume the rectangle has dimensions a x b\nDraw a radius to one of the verticies of the rectangle.\nThen drop a perpendicular from the center of the circle to the midpoint of one of the sides of the rectangle.\nThis forms a right triangle with sides a/2, b/2 and hypotenuse 5.\n(see picture: http://mudandmuck.com/str2/optimization.JPG )\n\nSo (a/2)^2 + (b/2)^2 = 5^2 = 25\nor a^2/4 + b^2/4 = 25\na^2 + b^2 = 100\nand b = √(100 - a^2).\n\nNow we want to maximize the area of the rectangle..\nThe area y = ab = a√(100 - a^2)\n\nTake the derivative with respect to a and set = 0.\n\ndy/da =  √(100 - a^2) +  a * (1/2) * (100 - a^2)^(-1/2) * (-2a)\n\ndy/da = √(100 - a^2) - a^2/√(100 - a^2)  = 0\n\nmultiply both sides by √(100 - a^2) to get:\n\n100 - a^2 - a^2 = 0\n2a^2 = 100\na^2 = 50\na = √50 = 5√2\nb = √(100 - a^2) = √50 = 5√2\n\nSo the rectangle is indeed a square of side length  5√2
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Just break the pentagon up into 5 triangles.  You can then add together the area of the 5 triangles to get the total area.
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asymptote at x=5 and y=2
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Which function, Subsitution, Linear COmbination, Graph?
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ANSWER : 25 square units.\n\n---------------------------------\n\nFirst off, draw a picture. \nSo, you have one point on the (0,0), one on the x-axis, one on the y-axis, and one on the diagonal line.\n\nLet's call the length of the horizontal line segment of the rectangle w.\n\nThe other side will then be (10-w).\n\nSo, the area of the rectangle is A = w*(10-w).\n\nIf you graph this, there is a maximum at the point (5,25), so 25 is the max area.\n\nTo verify this, take the derivative of A with respect to w, and set equal to zero.\nA' = 10 - 2w = 0\n2w = 10\nw = 5\nA = 5(5) = 25
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y=x^2 - 4x - 5\n\nFunction:\n\ny= 1(x-2)^2 - 9\n\nNow it's up to you to find out the rest. I not going to tell you everything.
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w1=width of 1st rect.\nL1=length of 1st rect\nA1=area of 1st rect\nw2=width of 2nd rect\nL2=length of 2nd rect\nA2=area of 2nd rect\nw1=x\nL1=4x\nA1=length*width\n=4*(square of x)\nw2=x+2\nL2=4x+5\nA2=(x+2)(4x+5)\n=x(4x+5)+2(4x+5)\n=4*(square of x)+5x+8x+10\n=4*(square of x)+13x+10\n\nAccording to the given condition\nA1+270=A2\n4*(square of x)+270=4*(square of x)+13x+10\n270=13x+10                //4*(square of x) gets cancelled\n13x=260\nx=20\ntherefore width of 1st rect=20cm\nlenght of 1st rect=20*4=80cm
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1. g(x)=2x-5\nso, g(f(x))=2f(x)-5\nor 2 (x^2+2x+3)-5\nor 2x^2+4x+1\n\n2. \na) put f(x)=y\nso y=2^x+1\n\nnow take independent values of x and find corresponding value of y to have 3 sets of locatins to find the graph\n\nIf x=0, then y=2 (2^0=1)\nif x=1, then y=3\nif x=2, then y=5\n\nb) now you have 3 sets of points (first one stands for x, position them in the graph paper against the function we have just found in (a) and see the domain..
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Divide the coefficient of the x term by 2:  -5/2\nThen square it:  (-5/2)^2 = 25/4\nAnswer: 25/4
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Intuitively I know the answer is:\n2.5 * -2.5 = -6.25\n\nBut let me show you how we can get the result a little more formally:\n\nStart with the two numbers:\nLet x be the smaller number\nLet x + 5 be the bigger number\n\nLet f(x) be the product:\nf(x) = x(x+5) \nf(x) = x² + 5x\n\nIf you graph it, this will be a parabola, and the vertex will be at the minimum point... but how do you figure out the vertex?\n\nWe need to massage the function so it is in vertex form... so first complete the square.  You do this by taking the coefficient on the x-term (5), dividing it in half (2.5) and squaring it (6.25).  Now add and subtract this from your equation:\n\nf(x) = x² + 5x + 6.25 - 6.25\n\nThe first part is now a perfect square:\nf(x) = (x + 2.5)(x + 25) - 6.25\n\nThis simplifies to:\nf(x) = (x + 2.5)² - 6.25\n\nCompare this to vertex form:\nf(x) = a(x - p)² + q\n\nHere's the equation in that form:\nf(x) = (x - (-2.5))² + (-6.25)\n\nSo:\na = 1\np = -2.5\nq = -6.25\n\nIf you graph this, you'll find your vertex (minimum point of the parabola) at the point (p, q).  This is the point (-2.5, -6.25).  \n\nSo your two numbers are:\n--> x = the smaller number = -2.5\n--> x + 5 = the larger number = 2.5\n\nThe product of -2.5 and 2.5 is -6.25, which is a minimum.
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(x+5)(x-5)\n(x)^2-(5)^2\n=x^2-25
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You're on the right track, but it gets involved. For example if the curve goes below the axis, then you've got to subtract the area in the rectangles that go below the axis. But for nice smooth curves that are positive, yes, adding up the area in the little rectangles is an approximation to the integral, and the approximation gets better as the rectangles get narrower.\n\nIn fact, by making the rectangles narrow enough you can get as close as you want to the actual integral.
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The graph of f(x) is a line, which means it can't have an asymptote.\n\ng(x) = log5(2x) should have a vertical asymptote x = 0.
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add 5 to both sides:    x^2 = 5\ntake the sqrt of both sides:  x= + or - sqrt of 5
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x  y\n0 0\n1 -1/5\n-5 1