can a math major please help me solve this intergral.  If possible can you include all of the steps thank you.
408042
Huh? To 3sqrx2 1? You'll need to give me more details, your question doesn't make sense otherwise.\n\nAnd use ^ as an exponent. I'm not sure if it's (8x^2 + 2x - 2) or (8x^2 + 2x^-2).\n\nAnyway, the indefinite integral of 8x^2 + 2x - 2 is (8x^3)/3 + x^2 - 2x. The indefinite integral of 8x^2 + 2x^-2 is (8x^3)/3 - 2/x. I can't get you any farther than that, because the rest of your question is incomprehensible.
774817
I can tell that nobody wants to just do this problem.  There's nothing interesting about it.  It's a typical problem that a stat student does to show understanding.  Get out your textbook and study up how to solve it.
823237
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
571352
I am assuming I means intergral, the way you find the areaunder a curve.\n\nA = Integral of (x^3+4x-3.6) = x^4/4 + 2x^2-3.6x\n\nusing the formula Integral of (x^n) = x^(n-1)/n\n\nOf course you need limits of integration. i.e. between x=2 & x=5 to get a numerical answer. i.e. 23.5\n\n\nUse these already proven exponent rules.\n\na^0 = 1,    so 5^0 = 1 \n\na^-n = 1/a^n,    so 5^-1 = 1/5^1 = 1/5 or 0.2\n\n\nAlso:\n5^0 = 1 because of the following exponent rule & division\n\na^m/a^n = a^(m-n) \n\nso 5^3/5^3 = 5^(3-3)= 5^0 \n\nhowever  \n\nso 5^3/5^3 = 125/125 = 1\n\nTherefore\n\n5^0 = 1\n\nThe exception to the rule is 0^0, because you can't divide by zero, although wriiten as a function its limit approcahes 0. But that's another class.\n\nYou can use the same process to come to the answer for\n\n5^-1 \n\n5^2/5^3 = 5^-1 = 25/125 = 1/5 or 0.2\n\nThese rules however are already proven.\n\na^0 = 1\n\na^-n = 1/a^n
28492
Let u=2x\ndu=2dx\n\nSubstitute in dx=du/2 and u=2x:\nintegral_of( csc2x dx ) = (1/2)integral_of( csc(u) du )\n\nMultiply numerator and denominator by: csc(u)+cot(u)\n==>(1/2)integral_of( [csc^2(u)+csc(u)cot(u)] du/\n[csc(u)+cot(u)] )\n==>(1/2)integral_of( (-1)[-csc^2(u)-csc(u)cot(u)] du/\n[csc(u)+cot(u)] )\n\nLet: v=csc(u)+cot(u)\ndv = [-csc(u)cot(u)-csc^2(u)] du\n\nSubstitute in the above:\n==>(1/2)integral_of(  (-1)dv/v  ) \n==>(-1/2)ln|v| + C\n==>(-1/2)ln|csc(u)+cot(u)| + C\n==>(-1/2)ln|csc(2x)+cot(2x)| + C
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
37738
try sosmath.com  they helped me before. if they don't have the answer you are looking for they have a list of links to tons of other math sites.  \n\nand there probably is someone on your campus that can help as well.  One on one help is always best...\n\nGood luck!
947331
(-1/4)(16/4) (16)\n-1 * 16\n-16\nReciprocal = -[1/16]
424650
Not really enough information to help, so try this site:\n\nhttp://www.blc.edu/fac/rbuelow/calc/ex4-6.html\n\nit's the second problem down the page
358834
KNOW YOUR RELATED RATES FOR CALC!! thats one thing that killed me and know your 2-D rotation around a line intergrals.
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
953940
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
811614
Answer:\nThe problem to solve is: \nRemember, with complex variables, keep like terms involving i together...\n \nAdding Multiplication Sign: y*x2-4x-5\nAdding Multiplication Sign: y*x*2-4x-5\nAdding Multiplication Sign: y*x*2-4*x-5\nPoly Parsing: y*x*2-4*x-5\nSplitting y*x*2-4*x-5 \ny*x*2-4*x - 5 \n[Left Side]Poly Parsing: y*x*2-4*x\nSplitting y*x*2-4*x \ny*x*2 - 4*x \n[Left Side]Poly Parsing: y*x*2\nMultiply Term y with Term x equals Term xy\nMultiply Term xy with Term 2 equals Term 2xy\nAddTerm: Term (2xy)\nPoly =  2xy\nRepace with Reduced Left Side \ny*x*2 - 4*x \n2xy - 4*x \n[Right Side]Poly Parsing: 4*x\nMultiply Term 4 with Term x equals Term 4x\nAddTerm: Term (4x)\nPoly =  4x\nRepace with Reduced Right Side \ny*x*2 - 4*x \n2xy - 4x \nBecause of the minus sign 4x becomes -4x\n\nPoly Adding 2xy with -4x...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\n... 2xy-4x\nPoly =  2xy-4x\nRepace with Reduced Left Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \n[Right Side]Poly Parsing: 5\nAddTerm: Term (5)\nPoly =  5\nRepace with Reduced Right Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \nBecause of the minus sign 5 becomes -5\n\nPoly Adding 2xy-4x with -5...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\nAddTerm: Term (-5)\n... 2xy-4x-5\nPoly =  2xy-4x-5\nThe problem to solve is: yx2-4x-5 \n2xy-4x-5
889632
The Ag in the product can not appear out of no where. I assume the equation should have been written:\n\nAl + AgNO3  >  Al(NO3)3 + Ag\n\nSo . . . Al + 3AgNO3  >  Al(NO3)3 + 3Ag\n\nGet an "A".
511709
Here is a link you may find helpful.\n\nhttp://www.math.buffalo.edu/mad/00.INDEXmad.html
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.