What is the integral from 0 to 2 of x^3sqrt(x^2+4) ?

The integral from 0 to 2 of x^3sqrt(x^2+4) is (64sqrt(2)+64)/(15)

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integral from 0 to 2 of x^3sqrt(x^2+4)

\int_{0}^{2} x^{3} x^{2} + 4 d x

\frac{64 2 + 64}{15}

Decimal Notation

10.30064 \dots

Solution steps

\int_{0}^{2} x^{3} x^{2} + 4 d x

Apply u-substitution

: \int_{2}^{22} (u^{2} - 4) u^{2} d u

= \int_{2}^{22} (u^{2} - 4) u^{2} d u

Expand

(u^{2} - 4) u^{2} : u^{4} - 4 u^{2}

= \int_{2}^{22} u^{4} - 4 u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{22} u^{4} d u - \int_{2}^{22} 4 u^{2} d u

\int_{2}^{22} u^{4} d u = \frac{128 2 - 32}{5}

\int_{2}^{22} 4 u^{2} d u = \frac{4 ( 16 2 - 8 )}{3}

= \frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3}

Simplify

\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3} : \frac{64 2 + 64}{15}

= \frac{64 2 + 64}{15}

\frac{\partial}{\partial y} (4 y - 3 x^{2})

\int \frac{x ^{4}}{( 3 + x ^{5} ) ^{2}} d x

\int \frac{1}{x \cdot ln ( x )} d x

\frac{d}{d x} (4 x + 10)

x \to 1 lim (3^{\frac{1}{x}})

What is the integral from 0 to 2 of x^3sqrt(x^2+4) ?
The integral from 0 to 2 of x^3sqrt(x^2+4) is (64sqrt(2)+64)/(15)