What is the integral from-1/2 to 0 of (2x+1)^{2020} ?

The integral from-1/2 to 0 of (2x+1)^{2020} is 1/4042

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integral from-1/2 to 0 of (2x+1)^{2020}

\int_{- \frac{1}{2}}^{0} (2 x + 1)^{2020} d x

\frac{1}{4042}

Decimal Notation

0.00024 \dots

Solution steps

\int_{- \frac{1}{2}}^{0} (2 x + 1)^{2020} d x

Apply u-substitution

: \int_{0}^{1} \frac{u ^{2020}}{2} d u

= \int_{0}^{1} \frac{u ^{2020}}{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_{0}^{1} u^{2020} d u

Apply the Power Rule

: [\frac{u ^{2021}}{2021}]_{0}^{1}

= \frac{1}{2} [\frac{u ^{2021}}{2021}]_{0}^{1}

Compute the boundaries:

\frac{1}{2021}

= \frac{1}{2} \cdot \frac{1}{2021}

Simplify

= \frac{1}{4042}

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What is the integral from-1/2 to 0 of (2x+1)^{2020} ?
The integral from-1/2 to 0 of (2x+1)^{2020} is 1/4042