What is the integral from-1 to 0 of 16x^5 ?

The integral from-1 to 0 of 16x^5 is -8/3

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integral from-1 to 0 of 16x^5

\int_{- 1}^{0} 16 x^{5} d x

- \frac{8}{3}

Decimal Notation

- 2.66666 \dots

Solution steps

\int_{- 1}^{0} 16 x^{5} d x

Take the constant out:

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

= 16 \cdot \int_{- 1}^{0} x^{5} d x

Apply the Power Rule

: [\frac{x ^{6}}{6}]_{- 1}^{0}

= 16 [\frac{x ^{6}}{6}]_{- 1}^{0}

Compute the boundaries:

- \frac{1}{6}

= 16 (- \frac{1}{6})

Simplify

= - \frac{8}{3}

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What is the integral from-1 to 0 of 16x^5 ?
The integral from-1 to 0 of 16x^5 is -8/3