What is the integral from 0 to 1 of x^2sqrt(4x+7) ?

The integral from 0 to 1 of x^2sqrt(4x+7) is (-343sqrt(7)+407sqrt(11))/(420)

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integral from 0 to 1 of x^2sqrt(4x+7)

\int_{0}^{1} x^{2} 4 x + 7 d x

\frac{- 343 7 + 407 11}{420}

Decimal Notation

1.05327 \dots

Solution steps

\int_{0}^{1} x^{2} 4 x + 7 d x

Apply u-substitution

: \int_{7}^{11} \frac{u ( u - 7 ) ^{2}}{64} d u

= \int_{7}^{11} \frac{u ( u - 7 ) ^{2}}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int_{7}^{11} u (u - 7)^{2} d u

Expand

u (u - 7)^{2} : u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 u

= \frac{1}{64} \cdot \int_{7}^{11} u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 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}{64} (\int_{7}^{11} u^{\frac{5}{2}} d u - \int_{7}^{11} 14 u^{\frac{3}{2}} d u + \int_{7}^{11} 49 u d u)

\int_{7}^{11} u^{\frac{5}{2}} d u = \frac{2662 11}{7} - 987

\int_{7}^{11} 14 u^{\frac{3}{2}} d u = \frac{14 ( 242 11 - 98 7 )}{5}

\int_{7}^{11} 49 u d u = \frac{49 ( 22 11 - 14 7 )}{3}

= \frac{1}{64} (\frac{2662 11}{7} - 987 - \frac{14 ( 242 11 - 98 7 )}{5} + \frac{49 ( 22 11 - 14 7 )}{3})

Simplify

\frac{1}{64} (\frac{2662 11}{7} - 987 - \frac{14 ( 242 11 - 98 7 )}{5} + \frac{49 ( 22 11 - 14 7 )}{3}) : \frac{- 343 7 + 407 11}{420}

= \frac{- 343 7 + 407 11}{420}

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What is the integral from 0 to 1 of x^2sqrt(4x+7) ?
The integral from 0 to 1 of x^2sqrt(4x+7) is (-343sqrt(7)+407sqrt(11))/(420)