Is finding the integral finding the sum of rectangles you construct beneath the curve?
814504
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.
406687
Here's the basic idea:  We usually estimate area under the curve using Riemann Sums, ie rectangles.  The height of these rectangles is most often based off the right endpoint, left endpoint, or the midpoint (though we could use any point - and we do when trying to prove the Fundamental Theorem of Calculus).\n\nI often ask students, "Given a Riemann Sum, how could we make it more accurate?"  Well, the answer I'm generally looking for is: use more rectangles... But, another common (and worthwhile) answer is: Average the Right Riemann sum and the Left Riemann Sum.  \n\nIn fact, this simple suggestion results in the trapezoid rule.  Try drawing some figures and convincing yourself that the average of the right and left triangles can be drawn as a trapezoid...\n\nSo, suppose we have n rectangles and an interval from a to b.  I will use the following notation:\n(delta)x = (b-a)/n\nx_0 = a\nx_1 = x_0 + (delta)x\n...\nx_n = b\n\nThen, we can write the Right Riemann Sum as follows:\nR = (delta)x*f(x_1) + (delta)x*f(x_2) + ... + (delta)x*f(x_n)\n=[f(x_1) + f(x_2) + ... + f(x_n)]*(delta)x\nWe can also write the Left Riemann Sum as follows: \nL = (delta)x*f(x_0) + (delta)x*f(x_1) + ... + (delta)x*f(x_n-1)\n=[f(x_0) + f(x_1) + ... + f(x_n-1)]*(delta)x\n\nTo get the trapezoid rule, we average them: T = (R+L)/2\n\nNotice that f(x_0) and f(x_n) are each used in only one formula each, but all the others are used in both sums.  So, basically, we get:\nT =(1/2)*[f(x_0) + 2*f(x_1) + 2*f(x_2)... + 2*f(x_n-1)+ f(x_n)]*(delta)x\n\nor, just to write it different we could say:\nT =[(1/2)*f(x_0) + f(x_1) + f(x_2)... + f(x_n-1)+ (1/2)*f(x_n)]*(delta)x
945543
A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
1109720
Simple answer: Any non-square rectangle can have a line drawn within it parallel to the short side that partitions it into a square and a smaller rectangle. If the long side of the rectangle is a and the short side is b, the new rectangle will have dimensions of b x (a-b). If this is not obvious to you, remember that one side of the rectangle is the side of the square (which has dimensions b x b), and the other is the part of the long side which is not occupied by the square (thus having a length equal to the length of the long side minus the length of the square). \n\nNow, some rectangles, when so partitioned, have the property that the new rectangle formed by the partition will have the same ratio between the lengths of its sides as the original rectangle. That is, a/b = b/(a-b). Such rectangles are called golden rectangles, and the ratio of the sides of a golden rectangle is called the golden ratio.\n\nThe golden ratio is easily shown to be both irrational and algebraic. Irrationality is proven by contradiction: note that all rational numbers can be represented in lowest terms - that is, as the ratio of two integers that have no common factors other than 1. However, the golden ratio cannot be represented in lowest terms, since if some ratio a/b were a representation in lowest terms, b/(a-b) would be a valid representation in even lower terms, which contradicts our statement that a/b is in lowest terms. Q.E.D.\n\nComputing the golden ratio is done by solving the equation we provided: a/b = b/(a-b). We arbitrarily choose b to be 1 - we could let b be anything, but letting b be 1 makes the ratio a/b much easier to compute, since it will simply be equal to a. Thus we have a=1/(a-1). Multiplying, we get a(a-1)=1 and from that a² - a = 1 Putting this in standard form, we have a² - a - 1 = 0, which shows that the golden ratio is algebraic, since our formula for computing it IS a polynomial - this stands in contrast to many other fundamental constants like pi and e, which cannot be expressed as a root of any polynomial of any degree (that is, they are non-algebraic).\n\nCompleting the square then yeilds:\n\na² - a - 1 = 0\na² - a + 1/4  = 5/4\n(a - 1/2)² = 5/4\na - 1/2 = ±√(5)/2\na = (1 ± √5)/2\n\nHowever, since the golden ratio is a ratio of lengths, and all lengths are positive, we know that it cannot be (1 - √5)/2, since that is negative. Thus the golden ratio is (1 + √5)/2 exactly.
406038
The oblique answer is correct. As for the slanted 3-D rectangle, I believe you are referring to a parallelepiped: http://mathworld.wolfram.com/Parallelepiped.html .
1165642
One way to reason is this.  Let n be any even integer greater than or equal to 4.  Divide the circle into n equal wedges (by drawing n/2 diameter lines with angles between successive lines equal to 360/n degrees).  Now, cut the wedges out and make a "rectangle" with them: n/2 wedges with the points downward and the arcs on top of adjacent pairs of wedges touching each other, then the other n/2 wedges with the points upward and adjacent arcs touching, filling the spaces of the first n/2 wedges.  What you end up with is almost a rectangle with top and bottom each made of n/2 360/n-degree arcs, and the lateral sides slanting slightly, made from the long edges of the wedges on each end.\n\nYou can estimate the area of this figure (which is the same as the area of the original circle of radius r) by pretending it IS a rectangle.  The length is the circumference of the circle, or 2*pi*r/2 (since half the circumference is the top, half is the bottom) (by definition of pi).  The width is the length of a wedge, or r.  So the area is about pi*r^2.  \n\nTo turn the estimate into an exact value, consider what happens as n goes to infinity, that is, it increases without bound.\n\nAs n goes to infinity, the slanted edges aproach being verticle and the number of arcs making up the top and bottom approaches infinity, and the arcs themselves approach straight horizontal lines.  Thus, the figure approaches a rectangle of exactly area pi*r^2.  However, for each n, you have the same circle of the same area, so the area of that circle is pi*r^2.
822332
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
1073502
Once again you will see that this is a rectangle.\nA = length times width\nA = 7*2\nA = 14
559279
Correct!\nOnly the second x's in the equation for asymptotes will be y.\nThis is an interesting result. Rectangular hyperbola is the locus of far corner of all rectangles' vertices having a fixed area A. And having the asymptotes as two adjascent sides of the rectangle.\nWith proper substitution, the general eqn can be transformed to ione.
660204
assume solution of the form sum(a sub n*x^n)\n\ndo two derivatives then plug into the equation, remembering for every derivative the indice of the sum is increased by 1.\n\nchange sin(x) into taylor polynomial then multiply by the power series expansion of y\n\nCompact all sums into one sum by reducing the indices.  \n\nFactor x terms out and set the coefficient of the x term equal to 0 to find the recursion formula for highest subscript of a sub n (I think it will be a sub (n+2) in this case).\n\nRe-enter this into your general solution from the beginning of the form sum(a sub n*x^n).\n\nViola!
246040
OK, so first you have to graph the functions, using your graphing calculator and you will see that the function y1=-x^2+4x+2 is above the function y2=x+2. And then we must find the intersection, (either by hand or by calculator)\nby hand:\ny1=y2\n-x^2+4x+2=x+2\nand solve for x\n3x-x^2=0\nx(3-x)=0\nx=0 and x=3\nand you'll get the points of intersection, (0,2) and (3,5).\nSince the cross section is perpendicular to the x-axis, you are forced to use dx.\nOk, then we can draw the squares in the area between the curve, and the squares have the base of top-bottom (i.e y1-y2) which is 3x-x^2. So the area of the one square is (3x-x^2)^2 right?\nOK then put it all in the cross section formula:\nintegral (from a to b)of A(x) dx\nso...\nintegral (from 0 to 3) of (3x-x^2)^2 dx \nThen for question b, you have to use equilateral triangle.. ok, so the height for equilateral triangle is (a/2)*(sqr)3 if the base is a. (Phytagorean Theorem)\nso your integral is\nintegral (from 0 to 3) of ((1/2)(sqr(3)/2)*(3x-x^2)^2) dx \nWell, I hope that's right, sorry if I made mistake.
1010092
Aaaaaargh you need CALCULUS aaaargh....\n\nIt took me 3 classses years ago to get a grip on that, but I've lost what I had.  You must be in high school or you would have been exposed to it by now - try looking up some calculus courses online, or possibly getting (yes I'm serious) one of those "Made Easy" textbooks, specifically "Calculus Made Easy".  Or look for basic calculus sites on the web, I found a few, but not specifically what you're looking for.  If I find something useful, I'll come back & post it.\n\nI think what you need to focus on are Limits, which are about breaking up the space under a curve into rectangles.\n\nI can say no more...\n\nFound Something!  It's Integrals.  Check out the link.\n\nHere's a better site:
942591
Chord length = R (see dimensions of rectangle ie flat part of slice)\nSo the size of the angle (θ) subtended at centre of circle is 60° as all sides of the triangle so formed = R.\nSo θ = π/3\n\nSo length of side of rectangle forming curved surface = length of Arc\n= Rθ\n= πR/3\n\nAlso area of segment at either end = ½R²(θ - sinθ)\n= ½R²(π/3 - sinπ/3)\n= ½R²(π/3 - √3/2)\n\nSo total surface area = Area of curved part + area of flat rectangle + 2 * area of end segment\n= πR/3 * 4R + R*4R + 2 * ½R²(π/3 - √3/2)\n= R²(5π/3 + 4 - √3/2)\n\nVolume = Area of segment * height\n= ½R²(π/3 - √3/2) * 4R\n= 2R³(π/3 - √3/2)
1089271
I don't know how these could be the sides of a rectangle since rectangles have two dimensions, not 3.  And if you mean a rectangular prism (box) none of these are invalid as dimensions.\n\nI think you mean triangle and the answers are 3,4 and 7 as well as 3,5 and 9 because that would result in triangles with no area.  \n\nThe sum of the shorter sides should always be more than the third side.  The only one that meets those criteria is #1.  \n\n#1:  2+3 > 5 --> good triangle\n#2:  3+4 = 7 --> bad triangle (flat)\n#3:  3+5 < 9 --> bad triangle (impossible)\n\nSo #2 and #3 are not the sides of a valid triangle.
18728
Again I assume you are talking about vector addition.  In this case you use trig the angle and magnitude given to find the x and y components for each vector.  Then just add all of the x componets together and all of the y components together to get the sum.  Then you can use the pathagorean theorem and arctan to find the magnitude and direction of the resultant vector.
123707
Let:\n\nL = length of the rectangle\nW = width of the rectangle\nP = perimeter of the rectangle\n\nGiven:\n\nW = x\nL = 2x - 4\nP = 2L + 2W from geometry\n\nSolution:\n\nP = 2L + 2W\n64 = 2[2x - 4] + 2x\n\n2[2x - 4] + 2x = 64\n\n[4x - 8] + 2x = 64\n\n4x - 8 + 2x = 64\n\n4x + 2x = 64 + 8\n\n6x = 72\n\nx  =  72/6\n\nx  =  12 cm.\n\nTherefore:\n\nW = width = x = 12 cm. ANS.\n\nL = length = 2x - 4 = 2(12) - 4 = 24 - 4 = 20 cm. ANS.
46001
you need to be more specific...\nbut technicaly the perimeter is the sum of all the sides...