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

User 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

User Jason Pan
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