Question Number 59679 by Forkum Michael Choungong last updated on 13/May/19
$${Evaluate} \\ $$$$\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\left(\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{{x}^{\mathrm{3}} }\right) \\ $$$$ \\ $$
Answered by Kunal12588 last updated on 13/May/19
![∫_0 ^3 (((x^2 +3x)/x^3 ))dx =∫_0 ^3 ((x^2 /x^3 )+((3x)/x^3 ))dx =∫_0 ^3 (1/x) dx + 3∫_0 ^3 (1/x^2 )dx =[ln(x)]_0 ^3 + 3[(x^(−2+1) /(−2+1))]_0 ^3 =ln(3)−ln(0)−3(3^(−1) −0^(−1) ) so ∫_1 ^3 (((x^2 +3x)/x^3 ))dx is undefined](Q59681.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\left(\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} }+\frac{\mathrm{3}{x}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{1}}{{x}}\:{dx}\:+\:\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$$=\left[{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{3}} \:+\:\mathrm{3}\left[\frac{{x}^{−\mathrm{2}+\mathrm{1}} }{−\mathrm{2}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$={ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{0}\right)−\mathrm{3}\left(\mathrm{3}^{−\mathrm{1}} −\mathrm{0}^{−\mathrm{1}} \right) \\ $$$${so}\:\int_{\mathrm{1}} ^{\mathrm{3}} \left(\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{{x}^{\mathrm{3}} }\right){dx}\:{is}\:{undefined} \\ $$
Answered by meme last updated on 13/May/19
![∫_0 ^3 ((x+3)/x^2 )=∫_0 ^3 (1/x)+∫_0 ^3 (1/x^2 )=[ln(x)]_0 ^3 −3[(1/x)]_0 ^3 impossible(ln0)](Q59692.png)
$$\int_{\mathrm{0}} ^{\mathrm{3}} \frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{1}}{{x}}+\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\left[{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{3}} −\mathrm{3}\left[\frac{\mathrm{1}}{{x}}\right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{impossible}\left({ln}\mathrm{0}\right) \\ $$
Answered by Joel122 last updated on 14/May/19
![Let f(x) = ((x^2 + 3x)/x^3 ) f(x) is undefined at x = 0, so it′s an improper integral I = ∫_0 ^3 ((x^2 + 3x)/x^3 ) dx = lim_(a→0^− ) [∫_a ^3 ((x^2 + 3x)/x^3 ) dx] [((∫_a ^3 ((x^2 + 3x)/x^3 ) dx = ∫_a ^3 (1/x) + (3/x^2 ) dx = [ln x − (3/x)]_a ^3 )),(( = ln ((3/a)) + (3/a) − 1)) ] I = lim_(a→0^− ) [ln ((3/a)) + (3/a) − 1] = ∞ ∴ The integral is divergent](Q59739.png)
$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}{{x}^{\mathrm{3}} } \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{undefined}\:\mathrm{at}\:{x}\:=\:\mathrm{0},\:\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{an}\:\mathrm{improper}\:\mathrm{integral} \\ $$$${I}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}{{x}^{\mathrm{3}} }\:{dx}\:=\:\underset{{a}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\left[\underset{{a}} {\overset{\mathrm{3}} {\int}}\:\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}{{x}^{\mathrm{3}} }\:{dx}\right] \\ $$$$\:\:\:\:\:\:\:\:\begin{bmatrix}{\underset{{a}} {\overset{\mathrm{3}} {\int}}\:\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}{{x}^{\mathrm{3}} }\:{dx}\:=\:\underset{{a}} {\overset{\mathrm{3}} {\int}}\:\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:{dx}\:=\:\left[\mathrm{ln}\:{x}\:−\:\frac{\mathrm{3}}{{x}}\right]_{{a}} ^{\mathrm{3}} }\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{ln}\:\left(\frac{\mathrm{3}}{{a}}\right)\:+\:\frac{\mathrm{3}}{{a}}\:−\:\mathrm{1}}\end{bmatrix} \\ $$$${I}\:=\:\underset{{a}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\left[\mathrm{ln}\:\left(\frac{\mathrm{3}}{{a}}\right)\:+\:\frac{\mathrm{3}}{{a}}\:−\:\mathrm{1}\right]\:=\:\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\therefore\:\mathrm{The}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{divergent} \\ $$