What is the integral from 0 to 3 of sqrt(1+81/64 x) ?

The integral from 0 to 3 of sqrt(1+81/64 x) is (307sqrt(307)-512)/(972)

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integral from 0 to 3 of sqrt(1+81/64 x)

\int_{0}^{3} 1 + \frac{81}{64} x d x

\frac{307 307 - 512}{972}

Decimal Notation

5.00727 \dots

Solution steps

\int_{0}^{3} 1 + \frac{81}{64} x d x

Apply u-substitution

: \int_{1}^{\frac{307}{64}} \frac{64 u}{81} d u

= \int_{1}^{\frac{307}{64}} \frac{64 u}{81} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{64}{81} \cdot \int_{1}^{\frac{307}{64}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{307}{64}}

= \frac{64}{81} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{307}{64}}

Compute the boundaries:

\frac{307 307}{768} - \frac{2}{3}

= \frac{64}{81} (\frac{307 307}{768} - \frac{2}{3})

Simplify

= \frac{307 307 - 512}{972}

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What is the integral from 0 to 3 of sqrt(1+81/64 x) ?
The integral from 0 to 3 of sqrt(1+81/64 x) is (307sqrt(307)-512)/(972)