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\int _(2.2)^(+\infty \:)(10.648)/(x^2)

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asked Jan 6, 2022 194k views

4 votes

$\int _(2.2)^(+\infty \:)(10.648)/(x^2)$

Mathematics
college

Ronni asked

by Ronni

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1 Answer

7 votes

Explanation:

$\int _(2.2)^(+\infty \:)(10.648)/(x^2)dx = 10.648 \int _(2.2)^(\infty \:)(dx)/(x^2)$

$= - (10.648)/(x) | _(2.2)^( \infty) = 4.84$

Jason Pan answered Jan 11, 2022

by Jason Pan

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\int _(2.2)^(+\infty \:)(10.648)/(x^2)

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