What is the integral from-1 to 0 of (2x+1)^{19} ?

The integral from-1 to 0 of (2x+1)^{19} is 0

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integral from-1 to 0 of (2x+1)^{19}

\int_{- 1}^{0} (2 x + 1)^{19} d x

0

Solution steps

\int_{- 1}^{0} (2 x + 1)^{19} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{u ^{19}}{2} d u

= \int_{- 1}^{1} \frac{u ^{19}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 1}^{1} u^{19} d u

Apply the Power Rule

: [\frac{u ^{20}}{20}]_{- 1}^{1}

= \frac{1}{2} [\frac{u ^{20}}{20}]_{- 1}^{1}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

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What is the integral from-1 to 0 of (2x+1)^{19} ?
The integral from-1 to 0 of (2x+1)^{19} is 0