What is the integral from 0 to 1 of xsqrt(x^2+8) ?

The integral from 0 to 1 of xsqrt(x^2+8) is 9-(16sqrt(2))/3

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integral from 0 to 1 of xsqrt(x^2+8)

\int_{0}^{1} x x^{2} + 8 d x

9 - \frac{16 2}{3}

Decimal Notation

1.45752 \dots

Solution steps

\int_{0}^{1} x x^{2} + 8 d x

Apply u-substitution

: \int_{22}^{3} u^{2} d u

= \int_{22}^{3} u^{2} d u

Apply the Power Rule

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

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

Compute the boundaries:

9 - \frac{16 2}{3}

= 9 - \frac{16 2}{3}

y = 3 x + 1

3 x sin (x)

\int x^{3} - x d x

\frac{\partial}{\partial x} (5 x^{2} + y^{2})

\int 3 cos (\frac{x}{2}) d x

What is the integral from 0 to 1 of xsqrt(x^2+8) ?
The integral from 0 to 1 of xsqrt(x^2+8) is 9-(16sqrt(2))/3