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

The integral from 0 to 4 of (x^3)/4 is 16

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

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

16

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

64

= \frac{1}{4} \cdot 64

Simplify

= 16

\int_{\frac{3 π}{2}}^{\frac{π}{2}} cos (x) d x

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

\int_{0}^{2} x + 6 - x^{3} d x

\int_{0}^{\frac{π}{4}} cos (2 x) d x

\int_{0}^{4} π (3 x^{2})^{2} d x

What is the integral from 0 to 4 of (x^3)/4 ?
The integral from 0 to 4 of (x^3)/4 is 16