What is the integral from 0 to 8 of sqrt(y+1) ?

The integral from 0 to 8 of sqrt(y+1) is 52/3

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integral from 0 to 8 of sqrt(y+1)

\int_{0}^{8} y + 1 d y

\frac{52}{3}

Decimal Notation

17.33333 \dots

Solution steps

\int_{0}^{8} y + 1 d y

Apply u-substitution

: \int_{1}^{9} u d u

= \int_{1}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{52}{3}

x \to 0 + lim (tan (\frac{4}{x}))

f (x) = x^{4} - x^{2}

t (\frac{d x}{d t}) + 3 x = 5 t^{2}

\int \frac{( x + 1 ) ^{2}}{x ^{2} - 1} d x

\frac{d}{d x} (12 x^{3} + 12 x^{2})

What is the integral from 0 to 8 of sqrt(y+1) ?
The integral from 0 to 8 of sqrt(y+1) is 52/3