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tony873004

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evaluate [tex]\int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx[/tex]

The back of the book gives 75/4 as the answer. I get 80/4. It has an asymptote at 1 since 0^(-1/5) DNE, and 0.00000001^(1/5) equals a number that grows larger as I add more 0's.

But any time I try to raise a negative number ^(1/5) I get an error. It just doesn't seem like 0-1 is in the Domain of this function. Even when I graph it, I see it shoot down from the x=1 asymptote, and then make a hard turn right as it approaches the x-axis. I don't see any Range in this function below x=1.

But somehow, the integral from 0 to 1 must equal -5/4 for my answer to match the back of the book's.

Here's my effort:

[tex]

\begin{array}{l}

\int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx \\

\\

\,\,\,\,\,\,\,\,\,\,u = x - 1,\,\,\frac{{du}}{{dx}} = 1,\,\,dx = du \\

\\

\int_*^* {u^{ - 1/5} } \,du = \mathop {\lim }\limits_{t \to 0^ + } \int_*^* {u^{ - 1/5} } \,du \\

\\

\mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \,\, = \,\,\mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \left. { = \mathop {\lim }\limits_{t \to 0^ + } \frac{{5\left( {x - 1} \right)^{4/5} }}{4}} \right|_t^{33} = \mathop {\lim }\limits_{t \to 0^ + } \left[ {\left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {t - 1} \right)^{4/5} }}{4}} \right)} \right] = \\

\\

\left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {0^ + - 1} \right)^{4/5} }}{4}} \right) = \frac{{5 \cdot 16}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} = \frac{{80}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} \\

\end{array}

[/tex]

But here's my problem.

If [tex]

\frac{{5\left( { - 1} \right)^{4/5} }}{4}=\frac{5}{4}[/tex] then I'd have the right answer.

The back of the book gives 75/4 as the answer. I get 80/4. It has an asymptote at 1 since 0^(-1/5) DNE, and 0.00000001^(1/5) equals a number that grows larger as I add more 0's.

But any time I try to raise a negative number ^(1/5) I get an error. It just doesn't seem like 0-1 is in the Domain of this function. Even when I graph it, I see it shoot down from the x=1 asymptote, and then make a hard turn right as it approaches the x-axis. I don't see any Range in this function below x=1.

But somehow, the integral from 0 to 1 must equal -5/4 for my answer to match the back of the book's.

Here's my effort:

[tex]

\begin{array}{l}

\int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx \\

\\

\,\,\,\,\,\,\,\,\,\,u = x - 1,\,\,\frac{{du}}{{dx}} = 1,\,\,dx = du \\

\\

\int_*^* {u^{ - 1/5} } \,du = \mathop {\lim }\limits_{t \to 0^ + } \int_*^* {u^{ - 1/5} } \,du \\

\\

\mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \,\, = \,\,\mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \left. { = \mathop {\lim }\limits_{t \to 0^ + } \frac{{5\left( {x - 1} \right)^{4/5} }}{4}} \right|_t^{33} = \mathop {\lim }\limits_{t \to 0^ + } \left[ {\left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {t - 1} \right)^{4/5} }}{4}} \right)} \right] = \\

\\

\left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {0^ + - 1} \right)^{4/5} }}{4}} \right) = \frac{{5 \cdot 16}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} = \frac{{80}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} \\

\end{array}

[/tex]

But here's my problem.

If [tex]

\frac{{5\left( { - 1} \right)^{4/5} }}{4}=\frac{5}{4}[/tex] then I'd have the right answer.

## Homework Statement

## Homework Equations

## The Attempt at a Solution

## Homework Statement

## Homework Equations

## The Attempt at a Solution

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