What is the integral from 0 to 4 of (x^3)/8 ?

The integral from 0 to 4 of (x^3)/8 is 8

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integral from 0 to 4 of (x^3)/8

\int_{0}^{4} \frac{x ^{3}}{8} d x

8

Solution steps

\int_{0}^{4} \frac{x ^{3}}{8} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{4} x^{3} d x

Apply the Power Rule

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

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

Compute the boundaries:

64

= \frac{1}{8} \cdot 64

Simplify

= 8

\int_{0}^{1} x (x^{2} + 5) d x

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

\int_{0}^{T} x d x

\int_{x}^{2} (1 + t^{2}) d t

\int_{0}^{2} (5 x^{3} - 6 x - 2) d x

What is the integral from 0 to 4 of (x^3)/8 ?
The integral from 0 to 4 of (x^3)/8 is 8