What is the integral from 0 to 4 of sqrt(y) ?

The integral from 0 to 4 of sqrt(y) is 16/3

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integral from 0 to 4 of sqrt(y)

\int_{0}^{4} y d y

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{4} y d y

Apply the Power Rule

: [\frac{2}{3} y^{\frac{3}{2}}]_{0}^{4}

= [\frac{2}{3} y^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

\frac{d}{d x} ((2 x + 5)^{4})

\frac{5}{s ^{2} + 2 s + 5}

\frac{d}{d x} (ln (1 + 4 x^{2}))

\int_{- 1}^{3} - x^{2} d x

\frac{d}{d x} (\frac{6}{5 x ^{3}})

What is the integral from 0 to 4 of sqrt(y) ?
The integral from 0 to 4 of sqrt(y) is 16/3