What is the integral from 0 to 6 of 9sqrt(4x+1) ?

The integral from 0 to 6 of 9sqrt(4x+1) is 186

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integral from 0 to 6 of 9sqrt(4x+1)

\int_{0}^{6} 9 4 x + 1 d x

186

Solution steps

\int_{0}^{6} 9 4 x + 1 d x

Take the constant out:

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

= 9 \cdot \int_{0}^{6} 4 x + 1 d x

Apply u-substitution

: \int_{1}^{25} \frac{u}{4} d u

= 9 \cdot \int_{1}^{25} \frac{u}{4} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int_{1}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

= 9 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

Simplify

9 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25} : \frac{9}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

= \frac{9}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

Compute the boundaries:

\frac{248}{3}

= \frac{9}{4} \cdot \frac{248}{3}

Simplify

= 186

36 y^{^{''}} - y = 0, y (- 5) = 1, y^{^{'}} (- 5) = - 1

y^{^{''}} - y^{^{'}} = 0

x \to 2 + lim (tan (\frac{π x}{4}))

\int_{1}^{\infty} 2^{- x} d x

f (x) = (x^{3} + 2)^{5}, at x = - 1

What is the integral from 0 to 6 of 9sqrt(4x+1) ?
The integral from 0 to 6 of 9sqrt(4x+1) is 186