What is the integral from 0 to 1 of sqrt(13/4+9/4 x) ?

The integral from 0 to 1 of sqrt(13/4+9/4 x) is (22sqrt(22)-13sqrt(13))/(27)

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integral from 0 to 1 of sqrt(13/4+9/4 x)

\int_{0}^{1} \frac{13}{4} + \frac{9}{4} x d x

\frac{22 22 - 13 13}{27}

Decimal Notation

2.08581 \dots

Solution steps

\int_{0}^{1} \frac{13}{4} + \frac{9}{4} x d x

Simplify

\frac{13}{4} + \frac{9}{4} x : \frac{1}{2} 13 + 9 x

= \int_{0}^{1} \frac{1}{2} 13 + 9 x d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} 13 + 9 x d x

Apply u-substitution

: \int_{13}^{22} \frac{u}{9} d u

= \frac{1}{2} \cdot \int_{13}^{22} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{9} \cdot \int_{13}^{22} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{22}

= \frac{1}{2} \cdot \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{22}

Simplify

\frac{1}{2} \cdot \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{22} : \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{22}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{22}

Compute the boundaries:

\frac{44 22 - 26 13}{3}

= \frac{1}{18} \cdot \frac{44 22 - 26 13}{3}

Simplify

= \frac{22 22 - 13 13}{27}

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What is the integral from 0 to 1 of sqrt(13/4+9/4 x) ?
The integral from 0 to 1 of sqrt(13/4+9/4 x) is (22sqrt(22)-13sqrt(13))/(27)