What is the integral from 0 to 6 of 3sqrt(4x+1) ?

The integral from 0 to 6 of 3sqrt(4x+1) is 62

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integral from 0 to 6 of 3sqrt(4x+1)

\int_{0}^{6} 3 4 x + 1 d x

62

Solution steps

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

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{4} \cdot \int_{1}^{25} u d u

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{248}{3}

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

Simplify

= 62

\int \frac{1}{x 6 x + 4} d x

y^{^{''}} = e^{3 t}

\int_{0}^{4} x^{3} e^{x} d x

x, 2 x - 12, 0, 8

\frac{d}{d x} (\frac{1}{8} x)

What is the integral from 0 to 6 of 3sqrt(4x+1) ?
The integral from 0 to 6 of 3sqrt(4x+1) is 62