What is the integral from 5 to 8 of sqrt(9/4 y-8) ?

The integral from 5 to 8 of sqrt(9/4 y-8) is (80sqrt(10)-13sqrt(13))/(27)

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integral from 5 to 8 of sqrt(9/4 y-8)

\int_{5}^{8} \frac{9}{4} y - 8 d y

\frac{80 10 - 13 13}{27}

Decimal Notation

7.63370 \dots

Solution steps

\int_{5}^{8} \frac{9}{4} y - 8 d y

Simplify

\frac{9}{4} y - 8 : \frac{1}{2} 9 y - 32

= \int_{5}^{8} \frac{1}{2} 9 y - 32 d y

Take the constant out:

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

= \frac{1}{2} \cdot \int_{5}^{8} 9 y - 32 d y

Apply u-substitution

: \int_{13}^{40} \frac{u}{9} d u

= \frac{1}{2} \cdot \int_{13}^{40} \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}^{40} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{40}

= \frac{1}{2} \cdot \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{40}

Simplify

\frac{1}{2} \cdot \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{40} : \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{40}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{40}

Compute the boundaries:

\frac{160 10 - 26 13}{3}

= \frac{1}{18} \cdot \frac{160 10 - 26 13}{3}

Simplify

= \frac{80 10 - 13 13}{27}

\int sin (2 x) sin (4 x) d x

\frac{\partial}{\partial x} (x 1 + y^{7})

sin (\frac{1}{x})

x \to 0 + lim (\frac{ln ( x )}{x})

What is the integral from 5 to 8 of sqrt(9/4 y-8) ?
The integral from 5 to 8 of sqrt(9/4 y-8) is (80sqrt(10)-13sqrt(13))/(27)