What is the integral from 0 to 1 of (15sqrt(5x+4))/2 ?

The integral from 0 to 1 of (15sqrt(5x+4))/2 is 19

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integral from 0 to 1 of (15sqrt(5x+4))/2

\int_{0}^{1} \frac{15 5 x + 4}{2} d x

19

Solution steps

\int_{0}^{1} \frac{15 5 x + 4}{2} d x

Take the constant out:

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

= \frac{15}{2} \cdot \int_{0}^{1} 5 x + 4 d x

Apply u-substitution

: \int_{4}^{9} \frac{u}{5} d u

= \frac{15}{2} \cdot \int_{4}^{9} \frac{u}{5} d u

Take the constant out:

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

= \frac{15}{2} \cdot \frac{1}{5} \cdot \int_{4}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

= \frac{15}{2} \cdot \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

Simplify

\frac{15}{2} \cdot \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9} : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

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

Compute the boundaries:

\frac{38}{3}

= \frac{3}{2} \cdot \frac{38}{3}

Simplify

= 19

\frac{d y}{d x} - 7 y = e^{x}

y^{^{'}} = \frac{11 x ^{2} - 1}{11 + 4 y}

h (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 5)^{3}

f (x) = \frac{7 x}{3 + x ^{2}}, at x = 2

x \to 1 lim (- \frac{1}{( x - 1 ) ^{2}})

What is the integral from 0 to 1 of (15sqrt(5x+4))/2 ?
The integral from 0 to 1 of (15sqrt(5x+4))/2 is 19