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Question Number 90512 by M±th+et£s last updated on 24/Apr/20

$${\int }_0^1\frac{1}{x}dx$$

Commented by Kunal12588 last updated on 24/Apr/20

$$\mathrm{\infty }?$$

Commented by jagoll last updated on 24/Apr/20

$$diverges$$

Answered by mahdi last updated on 24/Apr/20

$${=\ln \left(\mathrm{x}\right)]}_0^1=\ln \left(1\right)-\ln \left(0\right)=+\mathrm{\infty }$$

Commented by MJS last updated on 24/Apr/20

$$\mathrm{strictly}\mathrm{mathematical}\mathrm{language}$$$${\left[\ln x\right]}_0^1=\ln 1-\underset{x\to 0^{+}}{\lim ln}x...=+\mathrm{\infty }$$

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