What is the integral from-1 to 0 of x(x+1)^{100} ?

The integral from-1 to 0 of x(x+1)^{100} is -1/10302

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integral from-1 to 0 of x(x+1)^{100}

\int_{- 1}^{0} x (x + 1)^{100} d x

- \frac{1}{10302}

Decimal

- 0.00009 \dots

Solution steps

\int_{- 1}^{0} x (x + 1)^{100} d x

Apply u-substitution

: \int_{0}^{1} (u - 1) u^{100} d u

= \int_{0}^{1} (u - 1) u^{100} d u

Expand

(u - 1) u^{100} : u^{101} - u^{100}

= \int_{0}^{1} u^{101} - u^{100} 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^{101} d u - \int_{0}^{1} u^{100} d u

\int_{0}^{1} u^{101} d u = \frac{1}{102}

\int_{0}^{1} u^{100} d u = \frac{1}{101}

= \frac{1}{102} - \frac{1}{101}

Simplify

\frac{1}{102} - \frac{1}{101} : - \frac{1}{10302}

= - \frac{1}{10302}

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What is the integral from-1 to 0 of x(x+1)^{100} ?
The integral from-1 to 0 of x(x+1)^{100} is -1/10302