what is the derivation of the trapezoid rule?
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
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
1124961
To calculate perimeter of trapezoid you need to know the dimension of each side of trapeziod..\nthen.. sum of sides give you perimeter.
945908
Hi Dear Ruth ; \n\n{ as you know Sum ( angles of trapezoid) = 360\nif ' a ' & ' b ' & ' c ' & ' d ' are angles of trapezoid so \na + b + c + d = 360 } ....\n\n▪ Step 1 \na = 6x - 4   , b = 5x + 1 , c = 4x  , d = 3x + 3\nas , a + b + c + d = 360 so ;\n6x - 4 + 5x + 1 + 4x + 3x + 3 = 360\n[ 6x + 5x + 4x + 3x  ] + [ -4 + 1 + 3 ] = 360\n18x + 0 = 360 \nx = 360 / 14\nx = 20 \n\n▪ Step 2\nas x = 20 \na = 6x - 4  ; 6(20) - 4 = 116 \nb = 5x + 1 ; 5(20) + 1 = 101\nc = 4x ; 4(20) = 80 \nd = 3x + 3 ; 3(20) + 3 = 63\nso we have ;\na = 116 ; b = 101 , c = 80 , d = 63 \n\nGood Luck dear ♣
221721
Product rule:\n\n(fg)' = (f ')g + f(g ')\n\nSo,\n\n(-2x^2 * ln(x))' = (-4x * ln(x)) + (-2x^2 * 1/x)\n\nMultiply and clean up:\n\n-4xln(x) - 2x
454694
1\nDetails:\nLet y = lim x->0 of (sinx)^x\nTake the natural log of each side.\nln(y) = lim xln(sinx)\nWrite xln(sinx) as (ln(sinx))/(1/x)\nas x goes to 0, this goes to infinity over infinity.\nUse L'Hopital's rule \n(ln(sinx))/(1/x) becomes\n(cosx/sinx)/(-x^(-2))\n= (-x^2)(cosx)/(sinx)\nAs x goes to 0, this goes to 0/0.\nUse L'Hopital's rule  again\n(-x^2)(cosx)/(sinx) becomes\n[(x^2)(sinx) - 2x(cosx)]/(cosx)\nPlug in 0.\n[0-0]/1 = 0\nThis equals ln(y)\nln(y) = 0 means y = 1
766972
First off, this problem involves the chain rule. I think the person who said package rule really meant product rule, which is not requried here. \n\nThe chain rule allows us to find the deriviative of a function which we don't know how to differentiate in terms of functions which we do know how to differentiate. In this case, we know how to differentiate sin(3x), but not (sin(3x))^4 (I think it is less confusing to write it that way given the lack of mathematical notation available here).  The chain rule, in general, says that, if we have two functions, g(x) and h(x), we can find the deriviative of the compostie function g(h(x)) as follows:\n(g(h(x))' = g'(h(x)) * h'(x)\nSo, let us have hour h(x) function defined as:\nh(x) = sin(3x)\nIt's derivative is then:\nh'(x) = 3cos(3x)\nNow, lets define g(x) as:\ng(x) = x^4\ng'(x) = 4x^3\ng composed on h is then:\ng(h(x)) = g(sin(3x)) = (sin(3x))^4\ng' composed on h is:\ng'(h(x)) = 4(sin(3x))^3\nNow we have everything we need to use the chain rule, and find the derivative of g(h(x))\n(g(h(x))' = 4(sin(3x))^3 * 3cos(3x) \n             = 12*cos(3x)*(sin(3x))^3\nIn practice, one doesn't need to do this process so formally. Any time you have some base function you know how to differentiate raised  to some power, to take the derivative you simply bring the power out front, lower the power by one, and mulitply by the derivative of the base.
433476
the volume of a trapezoid-shaped box is the height of the box times the area of the trapezoid.\nwhere H is the height of the box, and h is the height of the trapezoid with bases b and B,\nV=1/2*(b+B)*h*H
1128734
I'm going to make a wild guess that you are trying to check an answer you got with the answer in the back of the book. I'm also going to assume that you were supposed to take a derivative of\n(x^3/(a-x))^(1/2).\nIn that case, you got the wrong answer because you didn't do the chain rule. You still need to multiply what you have by the derivative of the thing inside, i.e. the derivative of x^3/(a-x).\n\nIf I'm making a bad assumption and this didn't come from a differentiation problem, please ignore my answer.
539392
I'm thinking of a trapezoid with a base of 9 on top and 17 on the bottom, with the other two sides being 5.  If I were to drop a verticle line from one corner where the 9 base meets one of the legs to the other base, then I have a right triangle.  The hypotenuse would be 5 and the base would be (17-9)/2 = 4.  So the height would be sqrt(5^2-4^2) = 3.  So you have a trapezoid that has bases 9 and 17 and height 3, so the area is\n\nA = (17+9)3/2 = 39 (Area of trapezoid is (b1+b2)h/2, where b1 and b2 are the length of the bases and h is the height.)\n\nedit: Ooh, I just found a formula you can use if you know the length of the sides.  Let a be the length of the longer parallel side, c be the length of the shorter parallel side, and b and d be the legs.  Then\n\nA = (a+c)/(4(a-c))*\nsqrt((a+b-c+d)(a-b-c+d)*\n(a+b-c-d)(-a+b+c+d))\n\nSo in your case a = 17, c = 9, b = d = 5.\n\nA = (17+9)/(4(17-9))*\nsqrt((17+5-9+5)(17-5-9+5)*\n(17+5-9-5)(-17+5+9+5))\n= (26/32)sqrt(18*8*8*2)\n= (26/32)(48)\n= 39\n\nSorry, I have to line break sometimes because Yahoo likes to truncate.\n\nSo either way...
872821
You can prove it geometrically:\nhttp://www.mathwords.com/m/median_of_a_trapezoid.htm
350617
Actually trapezoids are Bottom Edge + Top Edge X Height /2.  Or:\n\nH(B+T)\n------\n  2
684619
Draw a horizontal staight line to be the base of the trapezoid.\nMark a point to be a corner of the trapezoid. Label it A.\nSet your compass to the length of the base.\nMeasure that distance from the first mark and make a second mark on the line. Lable it B\n\nWithout changing the width of the compass, put the point at A and swing an arc above the line. Move the point of the compass to B and swing and arc. Lable the point where the arcs cross C.\n\nDraw the lines AC and BC. You now have an equilateral triangle. The angles are all 60 degress.\n\nSet your compass to the distance of the leg length.\nPut the point at A and swing an arc across AC. Label the intersection D.\nMove the point to B and swing an arc across BC. Label the intersection E.\nDraw the line DE.\nDone.
893599
You may want to be more specific on this one... are you trying to find the largest possible area? You can inscribe lots of different trapezoids in a circle. \n\nAnd which theta are we talking about? The angle that the base makes with the circle?\n\nTry that one again,\n\nUMRmathmajor
595452
If you picture your trapezoid with the shorter parallel side and the two legs as sides 4, then the only unknown is the the long parallel side.  That side is at least 4, but with two extra parts on either side (if you drop perpendiculars from the short to the long side, you'd have a rectangle and two triangles).  So the long side is 4 + 2x, where x is the length of those two parts, which is also a leg of a triangle that we created in dropping those perpendiculars.\n\nThe height of the trapezoid is the same as the other leg of the triangle.  We don't know the length of either leg, but we do know that the hypotenuse is 4.  So through the Pythagorean theorem, we have\n\n4^2 = h^2 + x^2\nh^2 = 16 - x^2\nh = sqrt(16 - x^2)\n\nBack to the trapezoid - we know the area of a trapezoid is the average of the bases times the height, so we have\n\nA = (1/2)(4 + 4 + 2x)(sqrt(16 - x^2))\n= (4 + x)(sqrt(16 - x^2))\n= (4 + x)(16 - x^2)^(1/2)\n\nTo find the maximum area, take a derivative, set it to 0, and solve for x:\n\nA' = (4 + x)[(1/2)(16 - x^2)^(-1/2)](-2x) + [(16 - x^2)^(1/2)](1)\n= (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n\nIf we set this to 0, then\n0 = (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n-(-4x - x^2)(16 - x^2)^(-1/2) = (16 - x^2)^(1/2)\n4x + x^2 = 16 - x^2\n0 = 2x^2 + 4x - 16\n0 = x^2 + 2x - 8\n0 = (x + 4)(x - 2)\nx = -4 or 2\n\nSo your long side has to be 4 + 2(2) = 8 to maximize the area.
167445
The integral is defined over 6 units so 6/3 = 2 units.  Use your calculator to calculate f(0)=sin(0/3), f(2)=sin(2/3), f(4)=sin(4/3) and f(6)=sin(6/3).  Make sure your calculator is on "radians" not degrees when you do this.  The area of a trapezoid is the average of the heights times the base.\n\nWhen you use this method, the endpoints only get use once whereas the interior points get used twice.  You can now write:\n\nA = {[f(0)+f(2)]/2}2 + {[f(2)+f(4)]/2}2 +{[f(4)+f(6)]/2}2\n\nThe 2's cancel out, so you have f(0)+2f(2)+2f(4)+f(6).\n\nNow, put in your calculator values and add.
695595
Each question is a function of two variables... and it has two derivatives, relative to each variable.\n\nDerivative of F relative to Z1   is   uZ1^(u-1)*Z2^v\nDerivative of F relative to Z2   is   Z1^u*vZ2^(v-1)\n\nDerivative of MRTS relative to Z1 is  -uZ2/(vZ1^2)\n\nDerivative of MRTS relative to Z2 is  u/(vZ1)