MHB Evaluating Improper Integrals: Convergence or Divergence?

[brunette15]

I have the integral 1/(x^0.25 - 2) dx between 500 to 16, and am trying to find whether it converges or diverges.

I have sketched the graph and noticed that their is an asymptote at x=16 (hence why the integral is improper for these boundaries).

I am now trying to evaluate the limits to see if it converges or diverges but I am unsure how to approach this. Does anyone have any suggestions or hints? Thanks!

 

[Opalg]

I have the integral 1/(x^0.25 - 2) dx between 500 to 16, and am trying to find whether it converges or diverges.

I have sketched the graph and noticed that their is an asymptote at x=16 (hence why the integral is improper for these boundaries).

I am now trying to evaluate the limits to see if it converges or diverges but I am unsure how to approach this. Does anyone have any suggestions or hints? Thanks!

I would start by making the substitution $y = x^{1/4}$, so that $x=y^4$ and $dx = 4y^3dy.$

 

[Prove It]

I have the integral 1/(x^0.25 - 2) dx between 500 to 16, and am trying to find whether it converges or diverges.

I have sketched the graph and noticed that their is an asymptote at x=16 (hence why the integral is improper for these boundaries).

I am now trying to evaluate the limits to see if it converges or diverges but I am unsure how to approach this. Does anyone have any suggestions or hints? Thanks!

$\displaystyle \begin{align*} \int_{16}^{500}{ \frac{1}{\sqrt[4]{x} - 2} \, \mathrm{d}x} &= \int_{16}^{500}{ \frac{4\left( \sqrt[4]{x} \right) ^3}{4\left( \sqrt[4]{x} \right) ^3 \left( \sqrt[4]{x} - 2 \right) } \,\mathrm{d}x } \\ &= 4 \int_{16}^{500}{ \left[ \frac{\left( \sqrt[4]{x} \right) ^3}{ \sqrt[4]{x} - 2 } \right] \frac{1}{4 \left( \sqrt[4]{x} \right) ^3 } \, \mathrm{d}x } \end{align*}$

Now let $\displaystyle \begin{align*} u = \sqrt[4]{x} - 2 \implies \mathrm{d}u = \frac{1}{4 \left( \sqrt[4]{x} \right) ^3 } \, \mathrm{d}x \end{align*}$, and noting that $\displaystyle \begin{align*} u(500) = \sqrt[4]{500} - 2 \end{align*}$ and as $\displaystyle \begin{align*} x \to 16 , u \to 0 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 4 \int_{16}^{500}{ \left[ \frac{\left( \sqrt[4]{x} \right) ^3}{\sqrt[4]{x} - 2} \right] \frac{1}{4 \left( \sqrt[4]{x} \right) ^3 } \,\mathrm{d}x } &= 4 \int_0^{\sqrt[4]{500} - 2}{ \frac{\left( u + 2 \right) ^3}{u} \, \mathrm{d}u } \\ &= 4\int_0^{\sqrt[4]{500} - 2}{\frac{u^3 + 6u^2 + 12u + 8}{u}\,\mathrm{d}u} \\ &= 4 \lim_{b \to 0 }\int_b^{\sqrt[4]{500}-2}{ u^2 + 6u + 12 + \frac{8}{u}\,\mathrm{d}u } \end{align*}$

Can you finish?

 

[brunette15]

$\displaystyle \begin{align*} \int_{16}^{500}{ \frac{1}{\sqrt[4]{x} - 2} \, \mathrm{d}x} &= \int_{16}^{500}{ \frac{4\left( \sqrt[4]{x} \right) ^3}{4\left( \sqrt[4]{x} \right) ^3 \left( \sqrt[4]{x} - 2 \right) } \,\mathrm{d}x } \\ &= 4 \int_{16}^{500}{ \left[ \frac{\left( \sqrt[4]{x} \right) ^3}{ \sqrt[4]{x} - 2 } \right] \frac{1}{4 \left( \sqrt[4]{x} \right) ^3 } \, \mathrm{d}x } \end{align*}$

Now let $\displaystyle \begin{align*} u = \sqrt[4]{x} - 2 \implies \mathrm{d}u = \frac{1}{4 \left( \sqrt[4]{x} \right) ^3 } \, \mathrm{d}x \end{align*}$, and noting that $\displaystyle \begin{align*} u(500) = \sqrt[4]{500} - 2 \end{align*}$ and as $\displaystyle \begin{align*} x \to 16 , u \to 0 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 4 \int_{16}^{500}{ \left[ \frac{\left( \sqrt[4]{x} \right) ^3}{\sqrt[4]{x} - 2} \right] \frac{1}{4 \left( \sqrt[4]{x} \right) ^3 } \,\mathrm{d}x } &= 4 \int_0^{\sqrt[4]{500} - 2}{ \frac{\left( u + 2 \right) ^3}{u} \, \mathrm{d}u } \\ &= 4\int_0^{\sqrt[4]{500} - 2}{\frac{u^3 + 6u^2 + 12u + 8}{u}\,\mathrm{d}u} \\ &= 4 \lim_{b \to 0 }\int_b^{\sqrt[4]{500}-2}{ u^2 + 6u + 12 + \frac{8}{u}\,\mathrm{d}u } \end{align*}$

Can you finish?

Yes! Thankyou!