What is the integral from 0 to 1 of (1+9/4 x)^{1/2} ?

The integral from 0 to 1 of (1+9/4 x)^{1/2} is (13*13^{1/2}-8)/(27)

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integral from 0 to 1 of (1+9/4 x)^{1/2}

\int_{0}^{1} (1 + \frac{9}{4} x)^{\frac{1}{2}} d x

\frac{13 \cdot 1 3 ^{\frac{1}{2}} - 8}{27}

Decimal

1.43970 \dots

Solution steps

\int_{0}^{1} (1 + \frac{9}{4} x)^{\frac{1}{2}} d x

Simplify

(1 + \frac{9}{4} x)^{\frac{1}{2}} : \frac{1}{2} (4 + 9 x)^{\frac{1}{2}}

= \int_{0}^{1} \frac{1}{2} (4 + 9 x)^{\frac{1}{2}} 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} (4 + 9 x)^{\frac{1}{2}} d x

Apply u-substitution

: \int_{4}^{13} \frac{u ^{\frac{1}{2}}}{9} d u

= \frac{1}{2} \cdot \int_{4}^{13} \frac{u ^{\frac{1}{2}}}{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_{4}^{13} u^{\frac{1}{2}} d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

= \frac{1}{2} \cdot \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Simplify

\frac{1}{2} \cdot \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13} : \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Compute the boundaries:

\frac{26 \cdot 1 3 ^{\frac{1}{2}}}{3} - \frac{16}{3}

= \frac{1}{18} (\frac{26 \cdot 1 3 ^{\frac{1}{2}}}{3} - \frac{16}{3})

Simplify

= \frac{13 \cdot 1 3 ^{\frac{1}{2}} - 8}{27}

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What is the integral from 0 to 1 of (1+9/4 x)^{1/2} ?
The integral from 0 to 1 of (1+9/4 x)^{1/2} is (13*13^{1/2}-8)/(27)