What is the integral from 0 to 1 of x^3sqrt(2x+7) ?

The integral from 0 to 1 of x^3sqrt(2x+7) is (2(343sqrt(7)-891))/(45)

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integral from 0 to 1 of x^3sqrt(2x+7)

\int_{0}^{1} x^{3} 2 x + 7 d x

\frac{2 ( 343 7 - 891 )}{45}

Decimal Notation

0.73300 \dots

Solution steps

\int_{0}^{1} x^{3} 2 x + 7 d x

Apply u-substitution

: \int_{7}^{9} \frac{u ( u - 7 ) ^{3}}{16} d u

= \int_{7}^{9} \frac{u ( u - 7 ) ^{3}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{7}^{9} u (u - 7)^{3} d u

Expand

u (u - 7)^{3} : u^{\frac{7}{2}} - 21 u^{\frac{5}{2}} + 147 u^{\frac{3}{2}} - 343 u

= \frac{1}{16} \cdot \int_{7}^{9} u^{\frac{7}{2}} - 21 u^{\frac{5}{2}} + 147 u^{\frac{3}{2}} - 343 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

= \frac{1}{16} (\int_{7}^{9} u^{\frac{7}{2}} d u - \int_{7}^{9} 21 u^{\frac{5}{2}} d u + \int_{7}^{9} 147 u^{\frac{3}{2}} d u - \int_{7}^{9} 343 u d u)

\int_{7}^{9} u^{\frac{7}{2}} d u = 4374 - \frac{4802 7}{9}

\int_{7}^{9} 21 u^{\frac{5}{2}} d u = 21 (\frac{4374}{7} - 987)

\int_{7}^{9} 147 u^{\frac{3}{2}} d u = \frac{147 ( 486 - 98 7 )}{5}

\int_{7}^{9} 343 u d u = 343 (18 - \frac{14 7}{3})

= \frac{1}{16} (4374 - \frac{4802 7}{9} - 21 (\frac{4374}{7} - 987) + \frac{147 ( 486 - 98 7 )}{5} - 343 (18 - \frac{14 7}{3}))

Simplify

\frac{1}{16} (4374 - \frac{4802 7}{9} - 21 (\frac{4374}{7} - 987) + \frac{147 ( 486 - 98 7 )}{5} - 343 (18 - \frac{14 7}{3})) : \frac{2 ( 343 7 - 891 )}{45}

= \frac{2 ( 343 7 - 891 )}{45}

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What is the integral from 0 to 1 of x^3sqrt(2x+7) ?
The integral from 0 to 1 of x^3sqrt(2x+7) is (2(343sqrt(7)-891))/(45)