What is the integral from 0 to 4 of xsqrt(x^2+6) ?

The integral from 0 to 4 of xsqrt(x^2+6) is (22sqrt(22))/3-2sqrt(6)

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integral from 0 to 4 of xsqrt(x^2+6)

\int_{0}^{4} x x^{2} + 6 d x

\frac{22 22}{3} - 26

Decimal Notation

29.49740 \dots

Solution steps

\int_{0}^{4} x x^{2} + 6 d x

Apply u-substitution

: \int_{6}^{22} u^{2} d u

= \int_{6}^{22} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{6}^{22}

= [\frac{u ^{3}}{3}]_{6}^{22}

Compute the boundaries:

\frac{22 22}{3} - 26

= \frac{22 22}{3} - 26

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What is the integral from 0 to 4 of xsqrt(x^2+6) ?
The integral from 0 to 4 of xsqrt(x^2+6) is (22sqrt(22))/3-2sqrt(6)