What is the integral from-1 to 0 of (x+1)^{10}x ?

The integral from-1 to 0 of (x+1)^{10}x is -1/132

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integral from-1 to 0 of (x+1)^{10}x

\int_{- 1}^{0} (x + 1)^{10} x d x

- \frac{1}{132}

Decimal

- 0.00757 \dots

Solution steps

\int_{- 1}^{0} (x + 1)^{10} x d x

Apply u-substitution

= \int_{0}^{1} (u - 1) u^{10} d u

Expand

(u - 1) u^{10} : u^{11} - u^{10}

= \int_{0}^{1} u^{11} - u^{10} 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^{11} d u - \int_{0}^{1} u^{10} d u

\int_{0}^{1} u^{11} d u = \frac{1}{12}

\int_{0}^{1} u^{10} d u = \frac{1}{11}

= \frac{1}{12} - \frac{1}{11}

Simplify

\frac{1}{12} - \frac{1}{11} : - \frac{1}{132}

= - \frac{1}{132}

What is the integral from-1 to 0 of (x+1)^{10}x ?
The integral from-1 to 0 of (x+1)^{10}x is -1/132