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

The integral from-1 to 0 of x^2sqrt(1+x) is 16/105

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

\int_{- 1}^{0} x^{2} 1 + x d x

\frac{16}{105}

Decimal

0.15238 \dots

Solution steps

\int_{- 1}^{0} x^{2} 1 + x d x

Apply u-substitution

: \int_{0}^{1} (u - 1)^{2} u d u

= \int_{0}^{1} (u - 1)^{2} u d u

Expand

(u - 1)^{2} u : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

= \int_{0}^{1} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + 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_{0}^{1} u^{\frac{5}{2}} d u - \int_{0}^{1} 2 u^{\frac{3}{2}} d u + \int_{0}^{1} u d u

\int_{0}^{1} u^{\frac{5}{2}} d u = \frac{2}{7}

\int_{0}^{1} 2 u^{\frac{3}{2}} d u = \frac{4}{5}

\int_{0}^{1} u d u = \frac{2}{3}

= \frac{2}{7} - \frac{4}{5} + \frac{2}{3}

Simplify

\frac{2}{7} - \frac{4}{5} + \frac{2}{3} : \frac{16}{105}

= \frac{16}{105}

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What is the integral from-1 to 0 of x^2sqrt(1+x) ?
The integral from-1 to 0 of x^2sqrt(1+x) is 16/105