What is the integral from-1 to 0 of x^5sqrt(x^3+1) ?

The integral from-1 to 0 of x^5sqrt(x^3+1) is -4/45

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integral from-1 to 0 of x^5sqrt(x^3+1)

\int_{- 1}^{0} x^{5} x^{3} + 1 d x

- \frac{4}{45}

Decimal Notation

- 0.08888 \dots

Solution steps

\int_{- 1}^{0} x^{5} x^{3} + 1 d x

Apply u-substitution

: \int_{0}^{1} \frac{( u - 1 ) u}{3} d u

= \int_{0}^{1} \frac{( u - 1 ) u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} (u - 1) u d u

Expand

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

= \frac{1}{3} \cdot \int_{0}^{1} 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

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

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

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

= \frac{1}{3} (\frac{2}{5} - \frac{2}{3})

Simplify

\frac{1}{3} (\frac{2}{5} - \frac{2}{3}) : - \frac{4}{45}

= - \frac{4}{45}

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What is the integral from-1 to 0 of x^5sqrt(x^3+1) ?
The integral from-1 to 0 of x^5sqrt(x^3+1) is -4/45