What is the integral from 0.5 to 0.75 of 4x^3 ?

The integral from 0.5 to 0.75 of 4x^3 is 0.25390625

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integral from 0.5 to 0.75 of 4x^3

\int_{0.5}^{0.75} 4 x^{3} d x

0.25390625

Solution steps

\int_{0.5}^{0.75} 4 x^{3} d x

Take the constant out:

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

= 4 \cdot \int_{0.5}^{0.75} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0.5}^{0.75}

= 4 [\frac{x ^{4}}{4}]_{0.5}^{0.75}

Simplify

= 4 [0.25 x^{4}]_{0.5}^{0.75}

Compute the boundaries:

0.0634765625

= 4 \cdot 0.0634765625

Simplify

= 0.25390625

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What is the integral from 0.5 to 0.75 of 4x^3 ?
The integral from 0.5 to 0.75 of 4x^3 is 0.25390625