What is the integral from 0 to 2 of 7x^3sqrt(x^2+4) ?

The integral from 0 to 2 of 7x^3sqrt(x^2+4) is 7*(64sqrt(2)+64)/(15)

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integral from 0 to 2 of 7x^3sqrt(x^2+4)

\int_{0}^{2} 7 x^{3} x^{2} + 4 d x

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

Decimal Notation

72.10451 \dots

Solution steps

\int_{0}^{2} 7 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

= 7 \cdot \int_{0}^{2} x^{3} x^{2} + 4 d x

Apply u-substitution

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

= 7 \cdot \int_{2}^{22} (u^{2} - 4) u^{2} d u

Expand

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

= 7 \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

= 7 (\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}

= 7 (\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}

= 7 \cdot \frac{64 2 + 64}{15}

\frac{d}{d x} (\frac{x ^{2} + 6 x}{10 - x ^{2}})

y^{^{''}} - 7 y^{^{'}} + 10 y = - sin (3 t)

y = x^{2} + 2 x - 11, y = 3, [3, 5]

\frac{\partial}{\partial x} (4 x sin (7 x^{2} y))

f (x) = 3 x^{2} - 5 x

What is the integral from 0 to 2 of 7x^3sqrt(x^2+4) ?
The integral from 0 to 2 of 7x^3sqrt(x^2+4) is 7*(64sqrt(2)+64)/(15)