What is the integral from 0 to 2 of xsqrt(3x^2+4) ?

The integral from 0 to 2 of xsqrt(3x^2+4) is 56/9

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integral from 0 to 2 of xsqrt(3x^2+4)

\int_{0}^{2} x 3 x^{2} + 4 d x

\frac{56}{9}

Decimal

6.22222 \dots

Solution steps

\int_{0}^{2} x 3 x^{2} + 4 d x

Apply u-substitution

: \int_{4}^{16} \frac{u}{6} d u

= \int_{4}^{16} \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{4}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

Compute the boundaries:

\frac{112}{3}

= \frac{1}{6} \cdot \frac{112}{3}

Simplify

= \frac{56}{9}

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What is the integral from 0 to 2 of xsqrt(3x^2+4) ?
The integral from 0 to 2 of xsqrt(3x^2+4) is 56/9