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

The integral from 0 to 1 of xsqrt(4x^2+2) is (3sqrt(6)-sqrt(2))/6

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

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

\frac{3 6 - 2}{6}

Solution steps

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

Apply u-substitution

: \int_{2}^{6} \frac{u}{8} d u

= \int_{2}^{6} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{2}^{6} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{6}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{6}

Compute the boundaries:

46 - \frac{4 2}{3}

= \frac{1}{8} (46 - \frac{4 2}{3})

Simplify

= \frac{3 6 - 2}{6}

\int_{1}^{0} e^{- x} d x

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

\int_{- 4}^{4} x^{2} 64 - x^{3} d x

\int_{0}^{0.5} e^{x - 1} d x

\int_{0}^{0.5} 2 d x

What is the integral from 0 to 1 of xsqrt(4x^2+2) ?
The integral from 0 to 1 of xsqrt(4x^2+2) is (3sqrt(6)-sqrt(2))/6