What is the integral from 1 to 8 of sqrt(1+81/16 x) ?

The integral from 1 to 8 of sqrt(1+81/16 x) is (2656sqrt(83)-97sqrt(194))/(486sqrt(2))

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integral from 1 to 8 of sqrt(1+81/16 x)

\int_{1}^{8} 1 + \frac{81}{16} x d x

\frac{2656 83 - 97 194}{486 2}

Decimal Notation

33.24021 \dots

Solution steps

\int_{1}^{8} 1 + \frac{81}{16} x d x

Apply u-substitution

: \int_{\frac{97}{16}}^{\frac{83}{2}} \frac{16 u}{81} d u

= \int_{\frac{97}{16}}^{\frac{83}{2}} \frac{16 u}{81} d u

Take the constant out:

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

= \frac{16}{81} \cdot \int_{\frac{97}{16}}^{\frac{83}{2}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{97}{16}}^{\frac{83}{2}}

= \frac{16}{81} [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{97}{16}}^{\frac{83}{2}}

Compute the boundaries:

\frac{83 83}{3 2} - \frac{97 97}{96}

= \frac{16}{81} (\frac{83 83}{3 2} - \frac{97 97}{96})

Simplify

= \frac{2656 83 - 97 194}{486 2}

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What is the integral from 1 to 8 of sqrt(1+81/16 x) ?
The integral from 1 to 8 of sqrt(1+81/16 x) is (2656sqrt(83)-97sqrt(194))/(486sqrt(2))