What is the integral from-1 to 2 of 3xsqrt(4x^2-2) ?

The integral from-1 to 2 of 3xsqrt(4x^2-2) is (7sqrt(14)-sqrt(2))/2

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integral from-1 to 2 of 3xsqrt(4x^2-2)

\int_{- 1}^{2} 3 x 4 x^{2} - 2 d x

\frac{7 14 - 2}{2}

Decimal Notation

12.38869 \dots

Solution steps

\int_{- 1}^{2} 3 x 4 x^{2} - 2 d x

Take the constant out:

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

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

Apply u-substitution

: \int_{2}^{14} \frac{u}{8} d u

= 3 \cdot \int_{2}^{14} \frac{u}{8} d u

Take the constant out:

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

= 3 \cdot \frac{1}{8} \cdot \int_{2}^{14} u d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{14}

Simplify

3 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{14} : \frac{3}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{14}

= \frac{3}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{14}

Compute the boundaries:

\frac{28 14 - 4 2}{3}

= \frac{3}{8} \cdot \frac{28 14 - 4 2}{3}

Simplify

= \frac{7 14 - 2}{2}

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What is the integral from-1 to 2 of 3xsqrt(4x^2-2) ?
The integral from-1 to 2 of 3xsqrt(4x^2-2) is (7sqrt(14)-sqrt(2))/2