What is the integral from 0 to 1 of 7x^5sqrt(x^3+8) ?

The integral from 0 to 1 of 7x^5sqrt(x^3+8) is (7(512sqrt(2)-702))/(45)

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integral from 0 to 1 of 7x^5sqrt(x^3+8)

\int_{0}^{1} 7 x^{5} x^{3} + 8 d x

\frac{7 ( 512 2 - 702 )}{45}

Decimal

3.43425 \dots

Solution steps

\int_{0}^{1} 7 x^{5} x^{3} + 8 d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} x^{5} x^{3} + 8 d x

Apply u-substitution

: \int_{8}^{9} \frac{( u - 8 ) u}{3} d u

= 7 \cdot \int_{8}^{9} \frac{( u - 8 ) u}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int_{8}^{9} (u - 8) u d u

Expand

(u - 8) u : u^{\frac{3}{2}} - 8 u

= 7 \cdot \frac{1}{3} \cdot \int_{8}^{9} u^{\frac{3}{2}} - 8 u 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 \cdot \frac{1}{3} (\int_{8}^{9} u^{\frac{3}{2}} d u - \int_{8}^{9} 8 u d u)

\int_{8}^{9} u^{\frac{3}{2}} d u = \frac{486 - 256 2}{5}

\int_{8}^{9} 8 u d u = 8 (18 - \frac{32 2}{3})

= 7 \cdot \frac{1}{3} (\frac{486 - 256 2}{5} - 8 (18 - \frac{32 2}{3}))

Simplify

7 \cdot \frac{1}{3} (\frac{486 - 256 2}{5} - 8 (18 - \frac{32 2}{3})) : \frac{7 ( 512 2 - 702 )}{45}

= \frac{7 ( 512 2 - 702 )}{45}

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What is the integral from 0 to 1 of 7x^5sqrt(x^3+8) ?
The integral from 0 to 1 of 7x^5sqrt(x^3+8) is (7(512sqrt(2)-702))/(45)