What is the integral from 10 to y of (10)/(x^2) ?

The integral from 10 to y of (10)/(x^2) is -(10)/y+1

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integral from 10 to y of (10)/(x^2)

\int_{10}^{y} \frac{10}{x ^{2}} d x

- \frac{10}{y} + 1

Solution steps

\int_{10}^{y} \frac{10}{x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int_{10}^{y} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{10}^{y}

= 10 [- \frac{1}{x}]_{10}^{y}

Compute the boundaries:

- \frac{1}{y} + \frac{1}{10}

= 10 (- \frac{1}{y} + \frac{1}{10})

Simplify

= - \frac{10}{y} + 1

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What is the integral from 10 to y of (10)/(x^2) ?
The integral from 10 to y of (10)/(x^2) is -(10)/y+1