What is the integral from-1 to 0 of sqrt(y+1) ?

The integral from-1 to 0 of sqrt(y+1) is 2/3

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integral from-1 to 0 of sqrt(y+1)

\int_{- 1}^{0} y + 1 d y

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{- 1}^{0} y + 1 d y

Apply u-substitution

: \int_{0}^{1} u d u

= \int_{0}^{1} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

y^{^{'}} = 2 t, y (0) = 0

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\frac{\partial}{\partial x} (\frac{( 3 x - 2 y )}{x ^{2} + 3 y})

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What is the integral from-1 to 0 of sqrt(y+1) ?
The integral from-1 to 0 of sqrt(y+1) is 2/3