What is the integral from 1 to 2 of 2x^2sqrt(x^3+6) ?

The integral from 1 to 2 of 2x^2sqrt(x^3+6) is (2(28sqrt(14)-14sqrt(7)))/9

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integral from 1 to 2 of 2x^2sqrt(x^3+6)

\int_{1}^{2} 2 x^{2} x^{3} + 6 d x

\frac{2 ( 28 14 - 14 7 )}{9}

Decimal Notation

15.05019 \dots

Solution steps

\int_{1}^{2} 2 x^{2} x^{3} + 6 d x

Take the constant out:

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

= 2 \cdot \int_{1}^{2} x^{2} x^{3} + 6 d x

Apply u-substitution

: \int_{7}^{14} \frac{u}{3} d u

= 2 \cdot \int_{7}^{14} \frac{u}{3} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int_{7}^{14} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14}

= 2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14}

Simplify

2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14} : \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14}

= \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14}

Compute the boundaries:

\frac{28 14 - 14 7}{3}

= \frac{2}{3} \cdot \frac{28 14 - 14 7}{3}

Simplify

= \frac{2 ( 28 14 - 14 7 )}{9}

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What is the integral from 1 to 2 of 2x^2sqrt(x^3+6) ?
The integral from 1 to 2 of 2x^2sqrt(x^3+6) is (2(28sqrt(14)-14sqrt(7)))/9