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

The integral from 0 to 3 of sqrt(y+1) is 14/3

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integral from 0 to 3 of sqrt(y+1)

\int_{0}^{3} y + 1 d y

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

\int_{0}^{3} y + 1 d y

Apply u-substitution

: \int_{1}^{4} u d u

= \int_{1}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{14}{3}

f (x) = x^{3} + 3 x - 6, (1, - 2)

\frac{1}{s - 8}

\int \frac{8}{x ^{- 1} + 4} d x

\int_{0}^{1} \frac{2}{2 - x ^{2}} d x

\int \frac{1}{( x - 4 ) ( x + 3 )} d x

What is the integral from 0 to 3 of sqrt(y+1) ?
The integral from 0 to 3 of sqrt(y+1) is 14/3