the definite integral to within five decimals of accuracy. Round to five decimal places. Please help with the answer even if you dont want to show the work. I could swear i did it right, but i dont whats wrong
126101
First use that\ntan^-1(x)=x-(1/3)x^3+(1/5)x^5-+...\nand\nsin(x)=x-(1/6)x^3+(1/120)x^5-+...\n\nPlug in x^7 for x in these to get\ntan^-1(x^7)=x^7-(1/3)x^21+-..\nand\nsin(x^7)=x^7-(1/6)x^21+-...\n\nThen add the two and integrate term by term. When you plug in .4 and subtract what you get when you plug in 0 (which should be 0), you will get your answer.\n\nThe error from stopping at some term will be less than the value of the next term. Since .4^21 is about 4*10^(-9), you will only need the first term to get 5 decimal places of accuracy. I get .00041 for the answer, but check this.
456767
Those are the right numbers. As you suspected they have too many decimal places, however. To have them correct to 6 decimal places, look at the 7th decimal place. If this is 5 or more, round the 6th decimal up one, otherwise leave it the same and just drop all the numbers after it.\n\nSo for a) the answer would be 1.499975 and for b) it would be 1.499997.
184783
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
542448
It's ambiguous because you used no spaces or brackets.\nWas that "1/2 X+5" or "1/(2X+5)"?\n\nSince (int) (1/2 X+5)^6 dx is pretty easy as a Binomial Theorem expansion or substitution (Don't forget the constant of integration as person below did!), presume you meant the other possibility:\n\n(int) (1/(2X+5))^6 dx = (int) (2X+5)^-6 dx\nIntegrate by substituting (and possibly by parts): \nw = 2x+5, dw = 2dx\n\n(int) (2X+5)^-6 dx = 1/2 (int) w^-6 dw\n= (-1/2*7) w^-7 + C\n= (-1/14) w^-7 + C\n\n(Remember the constant of integration.)
478350
Please see the answer above and find out the missing factor in your calculation. You are in the right track. All the best!
659996
How does this work?  Is it okay to have a decimal in the middle?\n\nType 1   (Screen says 1)\n/ 3          (Screen says 0.3333333)\nx 3         (Screen says 0.9999999)\n+ 9        (Screen says 9.9999999)\n\n- Jack
473261
something is wrong in the way the problem is stated\nboth the answers are wrong, according to me.\nif you differentiate -19800-50.01x  you will get -50.01 and that cannot be equated to zero
462061
This is a very, very difficult (if not impossible) integral.  Which should give you a clue that it's a trick question.\n\nYou are supposed to notice that as x -> 0, x -> the lower bound.  The only reason that the integral might not be zero would be if the function (1 - tan(2t))^1/t approached infinity as t approaches zero.  So lets look at the function.\n\nAs t -> 0, (tan(2t)) -> 0.\nTherefore (1 - tan(2t)) -> 1\nWe know that as t -> 0, 1/t -> infinity.\nHowever, 1^x = 1 no matter what x is (even as x -> infinity)\n\nTherefore, the function approaches one as x -> 0.\nTherefore, the integral must be zero.
397740
Set u = Arccos(x) such that dx = sinu du.  Then the integral becomes the integral of (u^2)sinu du.  Now the integral of this is (by parts) = -(u^2)sinu + 2*(the integral of) (usinu du).  And the integral of usinu (by parts) = -ucosu+(the integral of)(cosu du).\n\nHope this helps,\n\nSteve
262630
First of all, if you round off, you loose the last significant digit. 1.5 rounds up to 2, not 2.0. \nWhen you write that a measurement is 2.0, you are saying that it lies between 1.95 and 2.05 and it was rounded to 2 significant places.\n\nHere is the convention for rounding:\nWhen round to a whole number, \nn.0 is already rounded.\nn.1, n.2, n.3, n.4 round down to n\nn.6, n.7, n.8, n.9 round up (n+1)\n\nTo avoid an upwards or downwards bias in your rounding, n.5 rounds up half the time, and down half the time, depending on whether n is even or odd. Always round so that n.5 in the direction that gives you an even number.\n\nSo 1.5 rounds to 2 and 2.5 rounds to 2.\n\nSimilarly -.5 rounds to 0, -1.5 and -2.5 round to -2.\n\nThe same rule applys whenever you round down to any number of significant digits.
189099
21.7-3.98=17.72, front end rounded=18\n\n14+9.76+25.8=49.76, front end rounded=18\n\nOn a separate note:\n\n8 1/3 (aka 8.333333333…) rounded to nearest 100th=8.33\n\ndividing 2.2 by 506; 2.2 divided by 506=.0043478260869565217391304347826087 rounded to nearest 100th=.00\n\n(if you were to switch this one around 506 by 2.2=230.00, which rounded to nearest 100th=.00 also).\n\n\nI still cannot reply to your message as I end up get this message from Yahoo Answers: “Oops\nYour email did not go through because the recipient's email address has not been confirmed.”
366293
you gotta integrate it, i forget what way it works but my guess is work is the integral of the force so do integral from 2 to 3 of 4x+3\nim sure you have to integrate just not sure which way check it up
47486
You need to do a Poisson integral. Rather than try to write things out, I refer you to eqaution (3) on the mathworld site.\n\n\nhttp://mathworld.wolfram.com/PoissonIntegral.html\n\nI hope that helps.
636222
err unless i'm misreading the question, then the answer is undefined.  the denominator is 0^3 which is 0.  you can't divide by 0.
347882
Integral: ∫ [2/{(e^2x)+4}] dx\n\nFirst consider the expression: 2/{(e^2x) + 4}\n\n= {e^(2x) + 4}/[2*{e^(2x) + 4}] - e^(2x)/[2*{e^(2x) + 4}]\n\n= 1/2 - (1/2)[e^(2x)/{e^(2x) + 4}]\n\n\nHence, ∫[2/{(e^2x) + 4}] dx\n\n=  ∫[(1/2) - (1/2)[e^(2x)/{e^(2x) + 4}]] dx\n\n=  ∫(1/2) dx - ∫(1/2)[e^(2x)/{e^(2x) + 4}] dx\n\n=  (1/2) x - (1/2)∫[e^(2x)/{e^(2x) + 4}] dx\n\nLet {e^(2x) + 4} = z. Then dz = 2e^(2x) dx\n\nHence the given integral becomes:\n(1/2) x - (1/4)∫(dz/z)\n\n= (1/2) x - (1/4)ln|z| + c where c = arbitrary constant of integration\n\nNow, substitute the value of 'z' in terms of 'x', thus the given integral becomes\n= (1/2) x - (1/4)ln|e^(2x) + 4| + c
650516
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
461179
It's undefined at the endpoint of the interval (pi/2).\nFirst go ahead and integrate it like you normally would.\nintegral of sin(x)/sqrt(cosx) dx\nUse substitution.  Let u = cos x, then du = - sinx dx\n= integral of -du/ sqrt(u) \n= integral -u^(-1/2) du\n= -u^(-1/2 +1) / (-1/2 + 1) \n= -u^(1/2) / (1/2)\n= -2u^(1/2)\n= -2 sqrt(u)\nReplace u with cosx\n= -2 sqrt(cos x)\nNow plug in pi/2\n- 2 sqrt(cos pi/2)\n= -sqrt(0)\n= 0\nPlug in pi/4\n-2 sqrt(cos pi/4)\n= -2sqrt( sqrt(2) / 2)\nNow subtract them\n0 - (-2sqrt(sqrt(2) /2))\n= 2sqrt( sqrt(2) / 2)