Evaluate
\frac{15}{32}=0.46875
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\int \frac{1}{x^{3}}\mathrm{d}x
Evaluate the indefinite integral first.
-\frac{1}{2x^{2}}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{3}}\mathrm{d}x with -\frac{1}{2x^{2}}.
-\frac{4^{-2}}{2}+\frac{1^{-2}}{2}
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
\frac{15}{32}
Simplify.
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