Improper Integrals

**jake210** · March 18th 2009, 03:51 PM

I'm having trouble with a couple of questions. I'm not sure about the last two, but I know that I need to use the limit to the lower bound for the first problem.

1) Integral of (3)/(x^2) dx bounded by -1 (lower) to 1 (upper)

2) Integral of (x^(-5/4)) dx bounded by 1 (lower) to infiniti (upper)

3) Integral of (1)/((8-x)^1/3) dx bounded by 0 (lower) to 8 (upper)

Thanks in advance for the help.

**Prove It** · March 18th 2009, 04:45 PM

Originally Posted by jake210

I'm having trouble with a couple of questions. I'm not sure about the last two, but I know that I need to use the limit to the lower bound for the first problem.

1) Integral of (3)/(x^2) dx bounded by -1 (lower) to 1 (upper)

2) Integral of (x^(-5/4)) dx bounded by 1 (lower) to infiniti (upper)

3) Integral of (1)/((8-x)^1/3) dx bounded by 0 (lower) to 8 (upper)

Thanks in advance for the help.

The first is improper because $x \neq 0$ .

So break it up into two integrals, namely

$\int_{-1}^0{\frac{3}{x^2}\,dx} + \int_0^1{\frac{3}{x^2}\,dx}$

$= \left[-\frac{3}{x}\right]_{-1}^0 + \left[-\frac{3}{x}\right]_0^1$

$= \left[\lim_{\varepsilon \to 0}\left(-\frac{3}{\varepsilon}\right) - 3\right] - \left[-3 - \lim_{\varepsilon \to 0}\left(-\frac{3}{\varepsilon}\right)\right]$ .

Can you go from here?

**jake210** · March 18th 2009, 04:50 PM

thanks for the help...i understand this problem now...how about the other two?

**Prove It** · March 18th 2009, 06:35 PM

Originally Posted by jake210

I'm having trouble with a couple of questions. I'm not sure about the last two, but I know that I need to use the limit to the lower bound for the first problem.

1) Integral of (3)/(x^2) dx bounded by -1 (lower) to 1 (upper)

2) Integral of (x^(-5/4)) dx bounded by 1 (lower) to infiniti (upper)

3) Integral of (1)/((8-x)^1/3) dx bounded by 0 (lower) to 8 (upper)

Thanks in advance for the help.

2. $\int_1^\infty{x^{-\frac{5}{4}}\,dx} = \left[-4x^{-\frac{1}{4}}\right]_1^\infty$

$= \lim_{\varepsilon \to \infty}\left(-4\varepsilon^{-\frac{1}{2}}\right) - (-4)$ .

Can you go from here?

3. $\frac{1}{(8 - x)^{\frac{1}{3}}} = (8 - x)^{-\frac{1}{3}}$ .

So evaluate $\int_0^8{(8 - x)^{-\frac{1}{3}}\,dx}$ .

Can you go from here?

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