What is the integral from 0 to 6 of 2xsqrt(4x+1) ?

The integral from 0 to 6 of 2xsqrt(4x+1) is 2188/15

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integral from 0 to 6 of 2xsqrt(4x+1)

\int_{0}^{6} 2 x 4 x + 1 d x

\frac{2188}{15}

Decimal

145.86666 \dots

Solution steps

\int_{0}^{6} 2 x 4 x + 1 d x

Take the constant out:

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

= 2 \cdot \int_{0}^{6} x 4 x + 1 d x

Apply u-substitution

: \int_{1}^{25} \frac{u ( u - 1 )}{16} d u

= 2 \cdot \int_{1}^{25} \frac{u ( u - 1 )}{16} d u

Take the constant out:

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

= 2 \cdot \frac{1}{16} \cdot \int_{1}^{25} u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= 2 \cdot \frac{1}{16} \cdot \int_{1}^{25} u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{16} (\int_{1}^{25} u^{\frac{3}{2}} d u - \int_{1}^{25} u d u)

\int_{1}^{25} u^{\frac{3}{2}} d u = \frac{6248}{5}

\int_{1}^{25} u d u = \frac{248}{3}

= 2 \cdot \frac{1}{16} (\frac{6248}{5} - \frac{248}{3})

Simplify

2 \cdot \frac{1}{16} (\frac{6248}{5} - \frac{248}{3}) : \frac{2188}{15}

= \frac{2188}{15}

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What is the integral from 0 to 6 of 2xsqrt(4x+1) ?
The integral from 0 to 6 of 2xsqrt(4x+1) is 2188/15