What is the integral from m to infinity of 1/(n^{6.63)} ?

The integral from m to infinity of 1/(n^{6.63)} is (0.17761…)/(m^{5.63)}

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integral from m to infinity of 1/(n^{6.63)}

\int_{m}^{\infty} \frac{1}{n ^{6.63}} d n

\frac{0.17761 \dots}{m ^{5.63}}

Solution steps

\int_{m}^{\infty} \frac{1}{n ^{6.63}} d n

Compute the indefinite integral:

\int \frac{1}{n ^{6.63}} d n = - \frac{0.17761 \dots}{n ^{5.63}} + C

Compute the boundaries:

\int_{m}^{\infty} \frac{1}{n ^{6.63}} d n = 0 - (- \frac{0.17761 \dots}{m ^{5.63}})

= 0 - (- \frac{0.17761 \dots}{m ^{5.63}})

Simplify

= \frac{0.17761 \dots}{m ^{5.63}}

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What is the integral from m to infinity of 1/(n^{6.63)} ?
The integral from m to infinity of 1/(n^{6.63)} is (0.17761…)/(m^{5.63)}