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Math Help - Evaluating integral (definite)

  1. #1
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    Evaluating integral (definite)

    Hi all
    Is my solution correct?

    Evaluate "integration sign" [(2e^x)-SQRT(x)] between limits b=10 and a=1.

    Integrating each tem separately:

    2e^x.dx,

    re-arranging: e^x.2.dx................(1)

    let u=x, du=1.dx, 2du=2dx

    substituting u=x and 2du=2dx back into (1) gives

    (e^u). 2du

    Integrated gives:

    (e^u). 2

    2e^x as u=x.

    Integrating SQRT(x)

    SQRT(x)= x^(1/2)

    Integrated gives:

    (x^3/2)/3/2

    Simplyfying

    [(2x^(3/2))/3]

    Combining both integrals

    [2e^x]-[(2x^(3/2))/3] and evaluating between limits of b=10 and a=1

    {[2e^x]-[(2x^(3/2))/3]}-{[2e^x]-[(2x^(3/2))/3]}

    {[2e^10]-[(2(10)^(3/2))/3]}-{[2e^(1)]-[(2(1)^(3/2))/3]}

    {44,052-21.1}-{5.4-0.66)

    44,026?

    thanks
    John
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  2. #2
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    Re: Evaluating integral (definite)

    Quote Originally Posted by celtic1234 View Post
    Hi all
    Is my solution correct?

    Evaluate "integration sign" [(2e^x)-SQRT(x)] between limits b=10 and a=1.

    Integrating each tem separately:

    2e^x.dx,

    re-arranging: e^x.2.dx................(1)

    let u=x, du=1.dx, 2du=2dx

    substituting u=x and 2du=2dx back into (1) gives

    (e^u). 2du

    Integrated gives:

    (e^u). 2

    2e^x as u=x.

    Integrating SQRT(x)

    SQRT(x)= x^(1/2)

    Integrated gives:

    (x^3/2)/3/2

    Simplyfying

    [(2x^(3/2))/3]

    Combining both integrals

    [2e^x]-[(2x^(3/2))/3] and evaluating between limits of b=10 and a=1

    {[2e^x]-[(2x^(3/2))/3]}-{[2e^x]-[(2x^(3/2))/3]}

    {[2e^10]-[(2(10)^(3/2))/3]}-{[2e^(1)]-[(2(1)^(3/2))/3]}

    {44,052-21.1}-{5.4-0.66)

    44,026?

    thanks
    John
    Why would you bother doing a u substitution for \displaystyle \int{2e^x\,dx}? Surely you know that \displaystyle \int{k\,f(x)\,dx} = k\int{f(x)\,dx} for constant \displaystyle k...

    It looks ok otherwise
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