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

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

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integral from-1 to 0 of x^2(x+1)^{1/2}

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

\frac{16}{105}

Decimal Notation

0.15238 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{2}{3} x^{2} (x + 1)^{\frac{3}{2}} - \int \frac{4}{3} x (x + 1)^{\frac{3}{2}} d x]_{- 1}^{0}

= [\frac{2}{3} x^{2} (x + 1)^{\frac{3}{2}} - \int \frac{4}{3} x (x + 1)^{\frac{3}{2}} d x]_{- 1}^{0}

\int \frac{4}{3} x (x + 1)^{\frac{3}{2}} d x = \frac{4}{3} (\frac{2}{5} x (x + 1)^{\frac{5}{2}} - \frac{4}{35} (x + 1)^{\frac{7}{2}})

= [\frac{2}{3} x^{2} (x + 1)^{\frac{3}{2}} - \frac{4}{3} (\frac{2}{5} x (x + 1)^{\frac{5}{2}} - \frac{4}{35} (x + 1)^{\frac{7}{2}})]_{- 1}^{0}

Compute the boundaries:

\frac{16}{105}

= \frac{16}{105}

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\int_{0}^{\infty} e^{- u^{2}} d u

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