Evaluate
\frac{3}{4096}=0.000732422
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\int _{-2}^{1}\frac{1}{16^{3}}\mathrm{d}x
Add 11 and 5 to get 16.
\int _{-2}^{1}\frac{1}{4096}\mathrm{d}x
Calculate 16 to the power of 3 and get 4096.
\int \frac{1}{4096}\mathrm{d}x
Evaluate the indefinite integral first.
\frac{x}{4096}
Find the integral of \frac{1}{4096} using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{1}{4096}\times 1-\frac{1}{4096}\left(-2\right)
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{3}{4096}
Simplify.
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