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

The integral from 0 to 2 of xsqrt(x^2+4) is (16sqrt(2)-8)/3

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

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

\frac{16 2 - 8}{3}

Decimal

4.87580 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{22} u^{2} d u

= \int_{2}^{22} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{2}^{22}

= [\frac{u ^{3}}{3}]_{2}^{22}

Compute the boundaries:

\frac{16 2}{3} - \frac{8}{3}

= \frac{16 2}{3} - \frac{8}{3}

Simplify

= \frac{16 2 - 8}{3}

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What is the integral from 0 to 2 of xsqrt(x^2+4) ?
The integral from 0 to 2 of xsqrt(x^2+4) is (16sqrt(2)-8)/3