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Thread: A few calc/integration questions.

  1. #1
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    A few calc/integration questions.

    Hi
    Just need some help with some calc/integration questions. Mainly just to check if i am doing it right.









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  2. #2
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    For (c) use substitution, let $\displaystyle u=\ln(x)$ , so $\displaystyle du=\frac{1}{x} dx$

    so the integral becomes $\displaystyle \int ^1 _0 u^{\frac{1}{n}} du$

    $\displaystyle \int ^1_0 u^{\frac{1}{n}} du= \frac{u^{\frac{1}{n}+1}}{\frac{1}{n}+1} |^1 _0 =\frac{1}{\frac{1}{n}+1}$
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  3. #3
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    Quote Originally Posted by akolman View Post
    For (c) use substitution, let $\displaystyle u=\ln(x)$ , so $\displaystyle du=\frac{1}{x} dx$

    so the integral becomes $\displaystyle \int ^1 _0 u^{\frac{1}{n}} du$

    $\displaystyle \int ^1_0 u^{\frac{1}{n}} du= \frac{u^{\frac{1}{n}+1}}{\frac{1}{n}+1} |^1 _0 =\frac{1}{\frac{1}{n}+1}$

    great thx, i got that one but i just made it n/n+1 in the end
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  4. #4
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    For 3(a) expand$\displaystyle (t^2+3t)^2$

    $\displaystyle (t^2+3t)^2=t^4+6t^3+9t^2$

    So,

    $\displaystyle \frac{(t^2+3t)^2}{t^3}=\frac{t^4+6t^3+9t^2}{t^3}
    = t + 6 + \frac{9}{t}$

    that way the integral is easier

    $\displaystyle \int \left( \frac{(t^2+3t)^2}{t^3} \right) dt= \int \left( t + 6 + \frac{9}{t} \right) dt$
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  5. #5
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    Quote Originally Posted by akolman View Post
    For 3(a) expand$\displaystyle (t^2+3t)^2$

    $\displaystyle (t^2+3t)^2=t^4+6t^3+9t^2$

    So,

    $\displaystyle \frac{(t^2+3t)^2}{t^3}=\frac{t^4+6t^3+9t^2}{t^3}
    = t + 6 + \frac{9}{t}$

    that way the integral is easier

    $\displaystyle \int \left( \frac{(t^2+3t)^2}{t^3} \right) dt= \int \left( t + 6 + \frac{9}{t} \right) dt$

    Got this one as well; thx

    I just need to know whether im right on the next 3.

    For 10 i got c
    For 5 i got c
    For 3 i got b

    are those correct?
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