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    Random Integration Question

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    Private Tutoring for 4u/3u/2u Mathematics, Sciences - Hurstville region

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    Integration reduction formula

    The reverse chain rule is integration by substitution, without actually doing the substitution, if that made any sense... lol
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    Integration reduction formula

    tan<sup>n-2</sup>x.sec<sup>2</sup>xdx = tan<sup>n-1</sup>x/(n-1) The derivative of tan x is sec<sup>2</sup>x, so if you use substitution of u= tan x, u get du = sec<sup>2</sup>xdx Then you are just integrating a power function of un-2 Hence, you get that result. It's the identical result for...
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    Recursion Formula

    cot(pi/2) = 0, as limit of tan(x) as x approaches pi/2 is +infinity
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    Some four unit question

    Haha, there's like a million different definitions for e.
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    Rectangular Hyperbola (mechanics)

    LOL... Guess you'll have to read the textbook by yourself.
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    Recursion Formula

    The same relation will still apply. You pull out a [cosec(x)]^2, so you'll be left with [cosec(x)]^(n-2) * d (cot(x)) Then integration by parts will solve all :)
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    4U Graphing question

    Rearrange the equation so y's are all on one side. x3 = -y3 + 3xy Then divide everything by x3 So: 1 = (-y3 + 3xy)/x3 = -(y/x)3 + 3y/x2 Then take the limits of x going to infinity. The 2nd term will become insignificant, so -(y/x)3 will tend to 1 as x tends to infinity. Hence, y = -x
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    Some four unit question

    No worries, it's all good :)
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    4U Graphing question

    You find you horizontal tangents and your vertical tangents. Then you mark those points in. Then, find your x and y intercepts, which should just be (0,0) Then, you find out what happens when x -> negative infinite. And the same for when x -> positive infinite. once you have these points...
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    Some four unit question

    Beat me to it :P
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    Some four unit question

    For part a, compare the areas under the graph. The upper rectangle (1/n) is greater than the actual area (integrate), and this area is greater than the lower rectangle (1/(n+1)) For part b, use the inequalities from part a. Using just the inequality on the LHS: We have 1/(n+1) < ln (1 + 1/n)...
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    Recursion Formula

    [cosec(x)]^2 dx = - d (cot(x)) Using that relationship, you can form your recursion formula, with integration by parts. Hope this helps.
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    Help with inverse function

    Sorry, I'm not capable of doing individual tutoring for any uni courses yet. Maybe wait one or two years :P. I'm assuming you are doing MATH1151 this semester. I take a support class for this course on Tuesdays 10-11 every week. Feel free to come along. The other support classes, I think are on...
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    Help with inverse function

    f'(x) = 3x<sup>2</sup> - 4x + 2 Discriminant = 16 - 24 < 0 Hence, no real roots exist for f'(x). So f'(x) > 0 for all values of x Therefore, f(x) is always an increasing function, so it is a one-to-one function.
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    Reverse Chain Rule ?

    oh, oops, i was thinking of reverse product rule. Yea, you pull a derivative out. It's basically substitution for the lazy.
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    Reverse Chain Rule ?

    The reverse chain rule is integration by parts. It's in the 4u course.
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    what happens if your raw mark in 4 unit is higher than ur raw mark in 3 unit

    If your mark in 4u is higher, then you're the best :P
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    old test questions i cant do. Conics

    b<sup>2</sup>=a<sup>2</sup>(e<sup>2</sup> - 1) so e<sup>2</sup> = (b<sup>2</sup>/a<sup>2</sup>) + 1 = 16/m<sup>2</sup> + 1 = (16 + m<sup>2</sup>)/m<sup>2</sup> Foci : (+ae, 0) So, the foci are: (+ sqrt(16 + m<sup>2</sup>), 0) Substitute this back into your line: 3x+y-6=0 Hence, 3 * sqrt (16...
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