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

29.49740 \dots

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Compute the boundaries:

\frac{22 22}{3} - 26

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

\int_{0}^{1} x^{2} (2 + 2 x^{3})^{5} d x

\int_{0}^{3} x^{2} e^{x^{3}} d x

\int_{0}^{π} - 2 cos^{2} (x) sin (x) d x

\int_{0}^{1} 3 cos (\frac{π x}{2}) d x

\int_{0}^{1} x^{2} (1 + 2 x^{3})^{3} d x

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)