What is the integral from 1 to 6 of x^2sqrt(x+3) ?

The integral from 1 to 6 of x^2sqrt(x+3) is 6856/35

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integral from 1 to 6 of x^2sqrt(x+3)

\int_{1}^{6} x^{2} x + 3 d x

\frac{6856}{35}

Decimal

195.88571 \dots

Solution steps

\int_{1}^{6} x^{2} x + 3 d x

Apply u-substitution

: \int_{4}^{9} (u - 3)^{2} u d u

= \int_{4}^{9} (u - 3)^{2} u d u

Expand

(u - 3)^{2} u : u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u

= \int_{4}^{9} u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 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

= \int_{4}^{9} u^{\frac{5}{2}} d u - \int_{4}^{9} 6 u^{\frac{3}{2}} d u + \int_{4}^{9} 9 u d u

\int_{4}^{9} u^{\frac{5}{2}} d u = \frac{4118}{7}

\int_{4}^{9} 6 u^{\frac{3}{2}} d u = \frac{2532}{5}

\int_{4}^{9} 9 u d u = 114

= \frac{4118}{7} - \frac{2532}{5} + 114

Simplify

\frac{4118}{7} - \frac{2532}{5} + 114 : \frac{6856}{35}

= \frac{6856}{35}

\int_{- 3}^{3} x^{3} - 9 x d x

\int_{0}^{π} 4 e^{x} + 7 sin (x) d x

\int_{0}^{3} (9 - x^{2})^{3} d x

\int_{- 1}^{1} (x^{2} - \frac{1}{3}) d x

\int_{- 3}^{- 1} (x^{3} + 6 x^{2} + 9 x + 9) d x

What is the integral from 1 to 6 of x^2sqrt(x+3) ?
The integral from 1 to 6 of x^2sqrt(x+3) is 6856/35