What is the integral from 0 to zy of z ?

The integral from 0 to zy of z is (y^2z^2)/2

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integral from 0 to zy of z

\int_{0}^{zy} z d z

\frac{y ^{2} z ^{2}}{2}

Solution steps

\int_{0}^{zy} z d z

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{0}^{yz}

= [\frac{z ^{2}}{2}]_{0}^{yz}

Compute the boundaries:

\frac{y ^{2} z ^{2}}{2}

= \frac{y ^{2} z ^{2}}{2}

\int_{0}^{\frac{π}{2}} (x - sin (x)) d x

\int_{0}^{2} - x^{2} d x

\int_{- 2}^{4} 1 - x - 4 x^{3} + 2 x^{5} d x

\int_{0}^{y} x (y^{2} - x^{2})^{\frac{3}{2}} d x

\int_{0}^{\frac{π}{2}} 4 sin (\frac{x}{2}) d x

What is the integral from 0 to zy of z ?
The integral from 0 to zy of z is (y^2z^2)/2