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Question Number 17086 by Joel577 last updated on 30/Jun/17

$$\underset{-1}{\stackrel{2}{\int }}\frac{1}{x^2}dx$$

Commented by Joel577 last updated on 30/Jun/17

$$\mathrm{How}\mathrm{to}\mathrm{solve}\mathrm{the}\mathrm{problem}\mathrm{like}\mathrm{this}?$$$$\mathrm{Because}\mathrm{when}\mathrm{I}\mathrm{graph}\mathrm{the}\mathrm{function},$$$$\mathrm{it}\mathrm{is}\mathrm{undefined}\mathrm{for}x=0$$

Commented by Joel577 last updated on 30/Jun/17

$$\mathrm{Is}\mathrm{it}-\frac{3}{2}\mathrm{or}\mathrm{\infty }?$$

Commented by ajfour last updated on 30/Jun/17

$$\mathrm{i}\mathrm{think}\mathrm{it}\mathrm{is}\mathrm{\infty }.$$

Commented by ajfour last updated on 30/Jun/17

$${\int }_{-1}^2\frac{1}{{\mathrm{x}}^2}\mathrm{dx}=(-\frac{1}{\mathrm{x}}){\mid }_{-1}^{-\mathrm{h}}+(-\frac{1}{\mathrm{x}}){\mid }_{\mathrm{h}}^2$$$$=\frac{1}{\underset{\mathrm{h}\to 0}{\lim }}(\frac{1}{\mathrm{h}}-1-\frac{1}{2}+\frac{1}{\mathrm{h}})$$$$=\mathrm{\infty }.$$

Commented by Joel577 last updated on 01/Jul/17

$$thankyouverymuch$$

Commented by 1234Hello last updated on 04/Jul/17

$$\mathrm{But}\mathrm{it}\mathrm{can}\mathrm{also}\mathrm{be}\mathrm{checked}\mathrm{that}\int \frac{dx}{x^2}$$$$=\frac{-1}{x}\mathrm{and}\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{at}x=0.$$

Answered by prakash jain last updated on 30/Jun/17

$$\mathrm{It}\mathrm{is}\mathrm{an}\mathrm{improper}\mathrm{integral}.$$$$2\mathrm{types}\mathrm{of}\mathrm{improper}\mathrm{integral}$$$$1.{\int }_a^{b}f\left(x\right)dx\mathrm{and}\mathrm{either}aand/orb$$$$\mathrm{is}\mathrm{unbouned}(-\mathrm{\infty }\mathrm{or}+\mathrm{\infty })$$$$2.{\int }_a^{b}f\left(x\right)dx\mathrm{where}f\left(x\right)\mathrm{is}\mathrm{unbounded}$$$$\mathrm{at}\mathrm{one}\mathrm{or}\mathrm{more}\mathrm{value}\mathrm{in}[a,b].\mathrm{Then}$$$$\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{split}\mathrm{the}\mathrm{integral}.$$$${\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx$$$$c\in [a,b]\mathrm{and}f\left(c\right)\mathrm{is}\mathrm{umbounded}.$$$$\mathrm{in}\mathrm{question}f\left(x\right)=\frac{1}{x^2}$$$$f\left(x\right)\mathrm{is}\mathrm{unbouned}\mathrm{at}x=0$$$${\int }_{-1}^2\frac{1}{x^2}dx\mathrm{is}\mathrm{convergent}\mathrm{only}\mathrm{if}$$$${\int }_{-1}^0\frac{1}{x^2}dx,{\int }_0^2\frac{1}{x^2}dx\mathrm{are}\mathrm{convergent}.$$$$\mathrm{if}\mathrm{any}\mathrm{of}\mathrm{the}\mathrm{split}\mathrm{integral}\mathrm{does}$$$$\mathrm{not}\mathrm{converge}\mathrm{the}\mathrm{combined}$$$$\mathrm{integral}\mathrm{does}\mathrm{not}\mathrm{converge}.$$$${\int }_{-1}^0\frac{1}{x^2}dx\mathrm{does}\mathrm{not}\mathrm{converge}.$$$$\mathrm{so}{\int }_{-1}^2\frac{1}{x^2}dx\mathrm{does}\mathrm{not}\mathrm{converge}.$$$${\int }_{-1}^2\frac{1}{x^2}dx\mathrm{integral}\mathrm{is}\mathrm{divergent}.$$

Commented by Joel577 last updated on 01/Jul/17

$$thankyouverymuch$$

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