What is the integral from 0 to 2 of 3sqrt(4x+1) ?

The integral from 0 to 2 of 3sqrt(4x+1) is 13

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integral from 0 to 2 of 3sqrt(4x+1)

\int_{0}^{2} 3 4 x + 1 d x

13

Solution steps

\int_{0}^{2} 3 4 x + 1 d x

Take the constant out:

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

= 3 \cdot \int_{0}^{2} 4 x + 1 d x

Apply u-substitution

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

= 3 \cdot \int_{1}^{9} \frac{u}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int_{1}^{9} u d u

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{52}{3}

= \frac{3}{4} \cdot \frac{52}{3}

Simplify

= 13

(2 x + 3) d x + (2 y - 2) d y = 0

\frac{d}{d x} (e^{- \frac{4}{x}})

y^{^{'}} = \frac{6 x ^{4} + y ^{4}}{x y ^{3}}

\int \frac{7 x ^{2}}{5 - x ^{6}} d x

x = - 3, x = 1, y = e^{x}, y = 1 - e^{x}

What is the integral from 0 to 2 of 3sqrt(4x+1) ?
The integral from 0 to 2 of 3sqrt(4x+1) is 13