Show that the converges \(\int_{∞}^{1}\frac{1}{x^2} \,dx\) and diverges \(\int_{∞}^{1}\frac{1}{x}\,dx\)

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Solution:

If limit of  \(\int_{∞}^{1}\frac{1}{x^2} \,dx\) exists then it converges

= \(\int_{∞}^{1}\frac{1}{x^2} \,dx\)

= \(\int_{∞}^{1} x^{-2} \,dx\)

= \(\begin{bmatrix}\frac{x^{-1}}{-1}\end{bmatrix}^{\!1}_{∞}\)

= \(\begin{bmatrix}\frac{-1}{x}\end{bmatrix}^{\!1}_{∞}\)

= -1 – \(\frac{1}{∞}\)

= -1

Since the limit exists so it converges.

 

Second Part:

If limit of  \(\int_{∞}^{1}\frac{1}{x} \,dx\) doesn’t exists then it diverges

= \(\int_{∞}^{1}\frac{1}{x} \,dx\)

= \(\begin{bmatrix} log(x) \end{bmatrix}^{\!1}_{∞}\)

= log(1) – log(∞)

= 0 – ∞

= – ∞

Since the limit doesn’t exists so it diverges.

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