What is the integral from 0 to 1 of x^2sqrt(6x+9) ?

The integral from 0 to 1 of x^2sqrt(6x+9) is (25sqrt(15)-54)/(35)

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integral from 0 to 1 of x^2sqrt(6x+9)

\int_{0}^{1} x^{2} 6 x + 9 d x

\frac{25 15 - 54}{35}

Decimal

1.22355 \dots

Solution steps

\int_{0}^{1} x^{2} 6 x + 9 d x

Apply u-substitution

= \int_{9}^{15} \frac{u ( u - 9 ) ^{2}}{216} d u

Take the constant out:

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

= \frac{1}{216} \cdot \int_{9}^{15} u (u - 9)^{2} d u

Expand

u (u - 9)^{2} : u^{\frac{5}{2}} - 18 u^{\frac{3}{2}} + 81 u

= \frac{1}{216} \cdot \int_{9}^{15} u^{\frac{5}{2}} - 18 u^{\frac{3}{2}} + 81 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}{216} (\int_{9}^{15} u^{\frac{5}{2}} d u - \int_{9}^{15} 18 u^{\frac{3}{2}} d u + \int_{9}^{15} 81 u d u)

\int_{9}^{15} u^{\frac{5}{2}} d u = \frac{6750 15 - 4374}{7}

\int_{9}^{15} 18 u^{\frac{3}{2}} d u = 18 (9015 - \frac{486}{5})

\int_{9}^{15} 81 u d u = 81 (1015 - 18)

= \frac{1}{216} (\frac{6750 15 - 4374}{7} - 18 (9015 - \frac{486}{5}) + 81 (1015 - 18))

Simplify

\frac{1}{216} (\frac{6750 15 - 4374}{7} - 18 (9015 - \frac{486}{5}) + 81 (1015 - 18)) : \frac{25 15 - 54}{35}

= \frac{25 15 - 54}{35}

What is the integral from 0 to 1 of x^2sqrt(6x+9) ?
The integral from 0 to 1 of x^2sqrt(6x+9) is (25sqrt(15)-54)/(35)