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

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

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

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

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

Decimal

20.60128 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Expand

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

= 2 \cdot \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

= 2 (\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}

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

Join

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

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

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

t an g e n t f (x) = x^{2} + x, at x = - 1, x = 2

s l o p e (8.7) (- 6.7)

n = 1 \sum \infty \frac{3}{n ^{2} + 4}

t a y l or \frac{1}{1 - a x}, 0

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