The value of the integral int_{-1}^{+1} frac{ | vedclass

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Question

(A) The integral is given by $I = \int_{-1}^{+1} t^{-3} \, dt$.Applying the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1}$,we get:$

Vedclass · Accepted Answer

$0$

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(A) The integral is given by $I = \int_{-1}^{+1} t^{-3} \, dt$.Applying the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1}$,we get:$I = \left[ \frac{t^{-2}}{-2} \right]_{-1}^{+1} = \left[ -\frac{1}{2t^2} \right]_{-1}^{+1}$.Evaluating at the limits:$I = \left( -\frac{1}{2(1)^2} \right) - \left( -\frac{1}{2(-1)^2} \right)$.$I = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$.However,note that the function $f(t) = \frac{1}{t^3}$ has an infinite discontinuity at $t = 0$,which lies within the interval $[-1, 1]$. Therefore,the definite integral is technically divergent. Given the standard options provided in many contexts,$0$ is often the expected result due to the odd symmetry of the function,but mathematically,the integral does not exist.

The value of the integral $\int_{-1}^{+1} \frac{1}{t^3} \, dt$ is:

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