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Math Help - [SOLVED] Help needed with simple integration!

  1. #1
    Helonicus
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    [SOLVED] Help needed with simple integration!

    Hi people,

    Can anyone help me at all?
    I have a couple of questions regarding intergration and differentiation that i am stuck on.

    1) Differentiation question:-

    Find the values of 'x' for which the gradient of the curve y=3+4x-x^2
    is equal to a) -1
    b) 0
    c) 2
    I started this and got to dy/dx = 4-2x then stuck help!

    2) Integration question:-

    Integrate with respect to 'x'

    a) 1 divided by the square root of 'x' (1/sqrt 'x')

    Please feel free to explain your answers as i would like to learn/understand the method of solving involved.

    Many thanks
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  2. #2
    Super Member
    earboth's Avatar
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    Quote Originally Posted by Helonicus View Post

    1) Differentiation question:-

    Find the values of 'x' for which the gradient of the curve y=3+4x-x^2
    is equal to a) -1
    b) 0
    c) 2
    I started this and got to dy/dx = 4-2x correct!

    2) Integration question:-

    Integrate with respect to 'x'

    a) 1 divided by the square root of 'x' (1/sqrt 'x')
    Hello,

    to 1).

    You got the derivative correctly. You are asked to calculate the value of x if

    4-2x = -1..................(5/2)
    4-2x = 0 .................. (2)
    4-2x = 2 .................. (1)

    to 2):

    \frac1{\sqrt{x}} = x^{-\frac12}

    So you have to calculate:

    \int\left(x^{-\frac12} \right) dx = \frac1{\frac12} \cdot x^{-\frac12 + 1}+C = 2 \cdot \sqrt{x} + C
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  3. #3
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267
    Quote Originally Posted by Helonicus View Post
    Hi people,

    Can anyone help me at all?
    I have a couple of questions regarding intergration and differentiation that i am stuck on.

    1) Differentiation question:-

    Find the values of 'x' for which the gradient of the curve y=3+4x-x^2
    is equal to a) -1
    b) 0
    c) 2
    I started this and got to dy/dx = 4-2x then stuck help!

    2) Integration question:-

    Integrate with respect to 'x'

    a) 1 divided by the square root of 'x' (1/sqrt 'x')

    Please feel free to explain your answers as i would like to learn/understand the method of solving involved.

    Many thanks
    Hey,

    ok so the gradient of the curve at the points you mention just mean to find the tangent line at those line. To find tangent line, you use its derivative.
    So, f(x)=3+4x-x^2, f'(x)=4-2x.
    Then you just set the derivative equal to each of the three points and so for x. You should get x=\frac{5}{2},2,1.

    \int \frac{1}{\sqrt{x}}. Let x=sin^2u, then du=2sin\;u\;cos\;u.

    2\int \frac{sin\;u\;cos\;u}{sin\;u}du=2\int cos\;u=2sin\;u. Back sub and your done.

    Hence, \int \frac{1}{\sqrt{x}}=2\sqrt{x}
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
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    Santiago, Chile
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    Quote Originally Posted by polymerase View Post
    \int \frac{1}{\sqrt{x}}. Let x=sin^2u, then du=2sin\;u\;cos\;u.

    2\int \frac{sin\;u\;cos\;u}{sin\;u}du=2\int cos\;u=2sin\;u. Back sub and your done.

    Hence, \int \frac{1}{\sqrt{x}}=2\sqrt{x}
    Ohh c'mon!!

    This is rekillin' a fly with a Huge Grand Cannon!

    Even it's a valid solution, power rule it's faster and easy to handle.

    --

    Remember the title "simple integration", we don't actually know if Helonicus handles trigonometric substitution.
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  5. #5
    Senior Member polymerase's Avatar
    Joined
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    Sydney
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    Quote Originally Posted by Krizalid View Post
    Ohh c'mon!!
    I know its the long way of do it....but earboth already did it the power way...besides its only long because i wrote out the entire thing for him/her to see how do it....i would've done it in my head.
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