What is the integral from 0 to 2 of x\sqrt[3]{4+x^2} ?

The integral from 0 to 2 of x\sqrt[3]{4+x^2} is (12-3\sqrt[3]{4})/2

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integral from 0 to 2 of x\sqrt[3]{4+x^2}

Decimal Notation

3.61889 \dots

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{4}^{8}

= \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{4}^{8}

Compute the boundaries:

Simplify

\frac{\partial}{\partial x} (\frac{3}{3 x + 5 y})

6 x^{2} ln (x), 24 ln (x)

\frac{d y}{d x} = \frac{1}{2} x + 2

x^{2} y^{^{''}} + 3 x y^{^{'}} + 2 y = 0, y (1) = 0.2, y^{^{'}} (1) = 0.3

\frac{d}{d x} (cos (sin (tan (π) x)))

What is the integral from 0 to 2 of x\sqrt[3]{4+x^2} ?
The integral from 0 to 2 of x\sqrt[3]{4+x^2} is (12-3\sqrt[3]{4})/2