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Math Help - Integration

  1. #1
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    Integration

    Could you explain the way to perform the following integrations, please?
    (see the attachment)

    Thank you!!!!!!!
    Attached Thumbnails Attached Thumbnails Integration-img055.jpg  
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  2. #2
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    In J you have a straight forward anti derivative x+e^x

    In K use u - substitution

    let u = x^4 +1 the anti-derviative is then simple
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  3. #3
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    Quote Originally Posted by student_forever View Post
    Could you explain the way to perform the following integrations, please?
    (see the attachment)

    Thank you!!!!!!!
    Please take the small time to type your questions. Perhaps even learn some basic latex, see here: http://www.mathhelpforum.com/math-help/latex-help/.

    \int_{-1}^1 1 + e^x \, dx
    Each of these terms in the integrand is a standard form. Refer to the tables here: Lists of integrals - Wikipedia, the free encyclopedia

    \int_0^4 x^3 \sqrt{x^4 + 1} \, dx
    Make the substitution u = x^4 + 1.
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  4. #4
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    for K:

    I have u=x^4+1 , du=4x^3, dx=1/4x^3
    when x=4, u=257; when x=0, u=1
    so now I have \intx^3\sqrt{u}*1/4x^3 correct?
    what do I do next?

    P.S. I am trying to write it in latex style, it doesn't work
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  5. #5
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    Quote Originally Posted by student_forever View Post
    for K:

    I have u=x^4+1 , du=4x^3, dx=1/4x^3
    when x=4, u=257; when x=0, u=1
    so now I have \intx^3\sqrt{u}*1/4x^3 correct?
    what do I do next?

    P.S. I am trying to write it in latex style, it doesn't work
    \frac{1}{4} \int_0^4 4x^3\sqrt{x^4+1} \, dx

    substitute ...

    \frac{1}{4} \int_1^{257} \sqrt{u} \, du

    now integrate.
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  6. #6
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    Sorry, guys, I just dont get it
    the answer should be 1/6[(257)^3/2-1)~686.50
    PLEASE, can you show me the solution step by step??
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  7. #7
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    What's the derivative of u^n? It's n*u^(n-1)*u', right? (just chain rule, bring the exponent out front, take one from that exponent, multiply by the derivative of the inside) Like, the derivative of (x^3 + 2)^2 is 2(x^3 + 2)(3x^2).

    When integrating, we just tak the opposite approach. That is to say, we raise n by 1, divide by n + 1, and then divide by the derivative of the inside.

    Look again at 2(x^3 + 2)(3x^2). Let's see if we can get back our original (x^3 + 2)^2. Let u = x^3 + 2. Since we're working backward, we know the form n*u^(n-1)*u' has an integral of u^n. Our (x^3 + 2) is being raised to the power of 1, right? So when does n-1=1? When n is 2. This means that so far, we're in the correct form n*u^(n-1). However, we haven't accounted for u'. If we are infact multiplying n*u^(n-1) by u', than we know our answer is just u^n. Lucky us, (3x^2) is u'. So it's in the correct form to convert to u^n, where n is 2 and u is x^3 + 2.

    Same concept with x^3(x^4 + 1)^(1/2). Let u=x^4 + 1. Our n is then 1/2 + 1 (or 3/2). Next, however, we don't have a 3/2 coefficient, do we? We have a coefficient of 1. So, multiply by 2/3 to account for that. There's just one other thing to check before we convert to u^n, our u'. Unfortunately again, we're multiplying by x^3, not 4x^3. But, like before, we just multiply by 1/4 to balance it out.

    Thus you get your answer of (2/3)(1/4)u^(3/2) or (1/6)u^(3/2)
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  8. #8
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    Quote Originally Posted by skeeter View Post
    \frac{1}{4} \int_0^4 4x^3\sqrt{x^4+1} \, dx

    substitute ...

    \frac{1}{4} \int_1^{257} \sqrt{u} \, du

    now integrate.
    \frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_1^{257}

    \frac{2}{12}\left[u^{\frac{3}{2}}\right]_1^{257}

    \frac{1}{6}\left[u^{\frac{3}{2}}\right]_1^{257}

    \frac{1}{6}\left[(257)^{\frac{3}{2}} - 1^{\frac{3}{2}}\right]

    \frac{1}{6}\left[(257)^{\frac{3}{2}} - 1\right]
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  9. #9
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    Skeeter, can you please write dow the solution to problem J?
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  10. #10
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    never mind, I got it!!

    Thanx to all of you!!!
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