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

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

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

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

- \frac{1}{y} + 1

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{y} + 1

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

\int_{0}^{2} \frac{1}{1 + x ^{3}} d x

\int_{0}^{\frac{π}{2}} ln (x) d x

\int_{0}^{6} (x^{2} + x + 1) d x

\int_{1}^{5} x^{3} + 8 x - 4 d x

\int_{3}^{4} 2 x^{2} d x

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