What is the integral from 1 to y of 2/(y^3) ?

The integral from 1 to y of 2/(y^3) is -1/(y^2)+1

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integral from 1 to y of 2/(y^3)

\int_{1}^{y} \frac{2}{y ^{3}} d y

- \frac{1}{y ^{2}} + 1

Solution steps

\int_{1}^{y} \frac{2}{y ^{3}} d y

Take the constant out:

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

= 2 \cdot \int_{1}^{y} \frac{1}{y ^{3}} d y

Apply the Power Rule

: [- \frac{1}{2 y ^{2}}]_{1}^{y}

= 2 [- \frac{1}{2 y ^{2}}]_{1}^{y}

Compute the boundaries:

- \frac{1}{2 y ^{2}} + \frac{1}{2}

= 2 (- \frac{1}{2 y ^{2}} + \frac{1}{2})

Simplify

= - \frac{1}{y ^{2}} + 1

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What is the integral from 1 to y of 2/(y^3) ?
The integral from 1 to y of 2/(y^3) is -1/(y^2)+1