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

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

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

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

\frac{52}{3}

Decimal

17.33333 \dots

Solution steps

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

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}

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