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Thread: Antiderivative question

  1. #1
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    Antiderivative question

    Let F(x) be an antiderivative of (ln x)^3/x. If F(1) = 0, then F(9) =

    a. .048
    b. .144
    c. 5.827
    d. 23. 308
    e. 1,640.250
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  2. #2
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    Quote Originally Posted by frozenflames
    Let F(x) be an antiderivative of (ln x)^3/x. If F(1) = 0, then F(9) =

    a. .048
    b. .144
    c. 5.827
    d. 23. 308
    e. 1,640.250
    Find,
    $\displaystyle \int \frac{(\ln x)^3}{x}$.
    Let $\displaystyle u=\ln x$
    then $\displaystyle du=\frac{1}{x}dx$
    Express the integral as,
    $\displaystyle \int (\ln x)^3\cdot \frac{1}{x}dx$
    Thus, we have after substitution,
    $\displaystyle \int u^3du=\frac{u^4}{4}+C$
    Thus, substituting back,
    $\displaystyle \frac{(\ln x)^4}{4}+C$
    But $\displaystyle F(1)=0$
    Thus,
    $\displaystyle \frac{(\ln 1)^4}{4}+C=0$
    Evaluate, the left part which gives,
    $\displaystyle 0+C=0$
    Thus, $\displaystyle C=0$
    Thus,$\displaystyle F(x)=\frac{(\ln x)^4}{4}$
    Thus, $\displaystyle F(9)=\frac{(\ln 9)^4}{4}\approx 5.83$

    $\displaystyle \cal{Q}.\cal{E}.\cal{D}$
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