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Math Help - HElp on definite integrals

  1. #1
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    HElp on definite integrals


    Answer key
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post

    Answer key

    Hello,

    I've done this integral in 2 steps:

    1. ∫(√(2x - x))dx = ∫(√(-((x - 2x + 1) - 1)))dx = ∫(√(1 - (x-1)))dx

    2. Substitute u = x-1 then du/dx = 1 that means du = dx

    You get:
    ∫(√(2x-x))dx = ∫(√(1 - u))du. Multiply the radicand by √(1 - u)/√(1 - u)

    ∫√(1 - u)du = ∫((1 - u)/√(1 - u))du. Split the integral into a sum:

    ∫(1 /√(1 - u))du - ∫((u)/√(1 - u))du. The first integral is arcsin(u).

    The second integral has to be done by partial integration. You'll get:

    ∫((u)/√(1 - u))du = -(u/2)*√(1 - u) + *arcsin(u)

    Put both integrals together(be careful: between the 2 integrals is a "-" sign!):

    ∫(√(1 - u))du = (u/2)*√(1 - u) + *arcsin(u)

    Now re-substitute:

    ∫(√(2x - x))dx = ((x-1)/2)*√(1 - (x-1)) + *arcsin(x-1)

    EB




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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post

    Answer key
    INT{0:2} sqrt(2x - x^2) dx

    Complete the square inside the radical to get:
    INT{0:2} sqrt(1 - (x - 1)^2) dx

    Let sinu = x - 1 <--> cosu du = dx
    INT{a:b} sqrt(1 - sin^2u)cosu du
    INT{a:b} sqrt(cos^2u)cosu du
    INT{a:b} cos^2u du

    Let cos2u = 2cos^2u - 1 --> cos^2u = 1/2(1 + cos2u)
    INT{a:b} 1/2(1 + cos2u) du
    1/2 INT{a:b} (1 + cos2u) du
    1/2 [u + 1/2sin2u] from {a:b}

    Let sin2u = 2sinu*cosu
    1/2 [u + 1/2(2sinu*cosu)] from {a:b}

    Since sinu = x - 1, cosu = sqrt(1 - (x - 1)^2), u = arcsin(x - 1)
    1/2 [arcsin(x - 1) + (x - 1)sqrt(1 - (x - 1)^2] from {0:2}
    1/2 [(arcsin(1) + (1)sqrt(1 - 1)) - (arcsin(-1) + (-1)sqrt(1 - 1))]
    1/2 [pi/2 + pi/2]
    1/2 (pi) = pi/2
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    INT{0:2} sqrt(2x - x^2) dx

    Complete the square inside the radical to get:
    INT{0:2} sqrt(1 - (x - 1)^2) dx

    Let sinu = x - 1 <--> cosu du = dx
    INT{a:b} sqrt(1 - sin^2u)cosu du
    INT{a:b} sqrt(cos^2u)cosu du
    INT{a:b} cos^2u du

    Let cos2u = 2cos^2u - 1 --> cos^2u = 1/2(1 + cos2u)
    INT{a:b} 1/2(1 + cos2u) du
    1/2 INT{a:b} (1 + cos2u) du
    1/2 [u + 1/2sin2u] from {a:b}

    Let sin2u = 2sinu*cosu
    1/2 [u + 1/2(2sinu*cosu)] from {a:b}

    Since sinu = x - 1, cosu = sqrt(1 - (x - 1)^2), u = arcsin(x - 1)
    1/2 [arcsin(x - 1) + (x - 1)sqrt(1 - (x - 1)^2] from {0:2}
    1/2 [(arcsin(1) + (1)sqrt(1 - 1)) - (arcsin(-1) + (-1)sqrt(1 - 1))]
    1/2 [pi/2 + pi/2]
    1/2 (pi) = pi/2
    oooh, that's nice. using a sin(u) substituion, i don't think i've ever seen that before. You're the man ecMathGeek!
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