What is the integral from 0 to 1 of 8x^3 ?

The integral from 0 to 1 of 8x^3 is 2

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integral from 0 to 1 of 8x^3

\int_{0}^{1} 8 x^{3} d x

2

Solution steps

\int_{0}^{1} 8 x^{3} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0}^{1}

= 8 [\frac{x ^{4}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= 8 \cdot \frac{1}{4}

Simplify

= 2

\int_{0}^{4 π} t^{2} sin (\frac{t}{2}) d t

\frac{d y}{d x} = 6 + 6 y + x + x y

n = 1 \sum \infty \frac{2 n}{n + 1}

\int 32 x^{3} - 15 x^{2} + 8 x d x

\int 2 x^{2} sin (3 x) d x

What is the integral from 0 to 1 of 8x^3 ?
The integral from 0 to 1 of 8x^3 is 2