What is the integral from 1 to 24 of sqrt(1+x) ?

The integral from 1 to 24 of sqrt(1+x) is (250-4sqrt(2))/3

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integral from 1 to 24 of sqrt(1+x)

\int_{1}^{24} 1 + x d x

\frac{250 - 4 2}{3}

Solution steps

\int_{1}^{24} 1 + x d x

Apply u-substitution

: \int_{2}^{25} u d u

= \int_{2}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{25}

= [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{25}

Compute the boundaries:

\frac{250}{3} - \frac{4 2}{3}

= \frac{250}{3} - \frac{4 2}{3}

Simplify

= \frac{250 - 4 2}{3}

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What is the integral from 1 to 24 of sqrt(1+x) ?
The integral from 1 to 24 of sqrt(1+x) is (250-4sqrt(2))/3