What is the integral from 0 to 2 of 6x^3sqrt(x^2+4) ?

The integral from 0 to 2 of 6x^3sqrt(x^2+4) is (2(64sqrt(2)+64))/5

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integral from 0 to 2 of 6x^3sqrt(x^2+4)

\int_{0}^{2} 6 x^{3} x^{2} + 4 d x

\frac{2 ( 64 2 + 64 )}{5}

Decimal Notation

61.80386 \dots

Solution steps

\int_{0}^{2} 6 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

= 6 \cdot \int_{0}^{2} x^{3} x^{2} + 4 d x

Apply u-substitution

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

= 6 \cdot \int_{2}^{22} (u^{2} - 4) u^{2} d u

Expand

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

= 6 \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

= 6 (\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}

= 6 (\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3})

Simplify

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

= \frac{2 ( 64 2 + 64 )}{5}

f (x) = - x^{2} + 2 x + 4, at x = 4

\int_{0}^{1} \frac{1}{e ^{3 x}} d x

\frac{d}{d x} (e^{- a x^{2}})

n = 0 \sum \infty \frac{e ^{n}}{π ^{n}}

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

What is the integral from 0 to 2 of 6x^3sqrt(x^2+4) ?
The integral from 0 to 2 of 6x^3sqrt(x^2+4) is (2(64sqrt(2)+64))/5