What is the integral from 0 to 4 of 0.125x^3 ?

The integral from 0 to 4 of 0.125x^3 is 8

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integral from 0 to 4 of 0.125x^3

\int_{0}^{4} 0.125 x^{3} d x

8

Solution steps

\int_{0}^{4} 0.125 x^{3} d x

Take the constant out:

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

= 0.125 \cdot \int_{0}^{4} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0}^{4}

= 0.125 [\frac{x ^{4}}{4}]_{0}^{4}

Simplify

= 0.125 [0.25 x^{4}]_{0}^{4}

Compute the boundaries:

64

= 0.125 \cdot 64

Simplify

= 8

e^{- x} arctan (2 e^{x})

\int (3 x^{2} + 4 x + 1) d x

n = 0 \sum \infty (- x^{2} - 8)^{n}

\int \frac{1}{2} e^{x^{2}} d x

(x^{2} - 4) \cdot \frac{d y}{d x} + 4 y = (x + 2)^{2}

What is the integral from 0 to 4 of 0.125x^3 ?
The integral from 0 to 4 of 0.125x^3 is 8