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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