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$\,\int\limits_{ - 1}^{ + 1} {\frac{1}{{{t^3}}}} \,dt$ સદીશનું મૂલ્ય  .  . . . થાય .

A

$0$

B

$1$

C

$-1$

D

$2$

Solution

$\,\,\int\limits_{ – 1}^{ + 1} {\frac{1}{{{t^3}}}} \,dt\,\,\,\, = \,\,\left[ {\frac{{{t^{ – 2}}}}{{ – 2}}} \right]_{ – 1}^{ + 1}$

$\, = \,\,\left[ { – \frac{1}{{2{t^2}}}} \right]_{ – 1}^{ + 1}$

$ = \,\,\left( { – \frac{1}{{2{{\left( 1 \right)}^2}}}} \right)\,\, – \,\left( { – \frac{1}{{2{{\left( { – 1} \right)}^2}}}} \right)\,\,$

$ = \,\, – \frac{1}{2}\,\, + \;\,\frac{1}{2}\,\, = \,\,0$

Standard 11
Physics

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