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Math Help - Some integral problems I'm having trouble with

  1. #1
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    Exclamation Some integral problems I'm having trouble with

    1. The integral: (sinˆ2 (3x) cos(3x) ) dx

    2. The integral: (x+2)(x-3)ˆ20 dx

    3. The integral from 0 to pi/4: sqrt(tan(x)) secˆ2 (x) ) dx

    4. The integral from 0 to 1: ( eˆ(cubed root of x) ) / 6xˆ(2/3) dx

    5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.
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  2. #2
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    Integral problems!

    1. The integral: (x+2)(x-3)ˆ20 dx

    2. The integral from 0 to 1: ( eˆ(cubed root of x) ) / 6xˆ(2/3) dx

    3. The integral from 0 to pi/4: sqrt(tan(x)) secˆ2 (x) ) dx

    4. The integral: (sinˆ2 (3x) cos(3x) ) dx

    5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.
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  3. #3
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    Quote Originally Posted by coolio View Post
    1. The integral: (x+2)(x-3)ˆ20 dx
    The integrand is the derivative of f(x)=(ax+b)(x-3)^21 for same a and b.
    Differentiating f(x) gives:

    f'(x) = (x-3)^20 [22*a*x + 21*(ax+b)]

    equating coefficients gives a=1/22, and b=47/(21*22)

    So by the fundamental theorem of calculus:

    Integral (x+2)(x-3)ˆ20 dx = (1/22)(x+47/21)(x-3)^21 +C

    RonL
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  4. #4
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    first 2 problems only

    Quote Originally Posted by coolio View Post
    1. The integral: (sinˆ2 (3x) cos(3x) ) dx

    2. The integral: (x+2)(x-3)ˆ20 dx

    3. The integral from 0 to pi/4: sqrt(tan(x)) secˆ2 (x) ) dx

    4. The integral from 0 to 1: ( eˆ(cubed root of x) ) / 6xˆ(2/3) dx

    5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.
    Hi,

    to 1.) Use substitution:

    u(x) = sin(3x) ===> du/dx = 3*cos(3x) ===> du = 3*cos(3x) * dx

    Your integral becomes:

    (1/3)∫(sin(3x) * 3*cos(3x) dx = (1/3) ∫u*du = 1/9*u + C

    resubstitute:

    ∫(sin(3x) * cos(3x) dx = 1/9*sin(3x) + C


    to 2.)
    Use partial integration:

    Code:
    ∫(x+2)(x-3)^20 dx = 1/21*(x+2)(x-3)^21 - ∫1*1/21*(x-3)^21 dx
    
    u  = x+2      v  = 1/21*(x-3)^21
    u' = 1        v' = (x-3)^20
    Thus your integral becomes:

    ∫(x+2)(x-3)^20 dx = 1/21*(x+2)(x-3)^21 - 1/462*(x-3)^22 + C

    EB
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  5. #5
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    Quote Originally Posted by coolio View Post
    ...
    5. Find the area that is enclosed by the curve y=2x-x3 , the line x=1, the line x=5, and the x axis.
    Hi,

    in the given interval there is a zero of the function. Therefore you have to split the interval into [1, √(2)] and ]√(2), 5].

    ∫ from 1 to √(2) (2x-x)dx = (x - 1/4x^4) from 1 to √(2) = (2-1) - (1-1/4) = 1/4

    ∫ from √(2) to 5 (2x-x)dx = (x - 1/4x^4) from √(2) to 5 = (22-625/4) - (2-1) = -529/4

    Add the absolute values. The area is 1/4 + 529/4 = 530/4 = 265/2 = 1325.5

    EB
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  6. #6
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    Hello, coolio!

    3) .(tan x)^ secx dx
    Let: u = tan x



    . . . . .e^(x^{1/3})
    4) .
    --------------- dx
    . . . . . .6x^{2/3}
    Let: u = x^{1/3}

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