What is the integral from 3 to 5 of (x^3)/8 ?

The integral from 3 to 5 of (x^3)/8 is 17

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integral from 3 to 5 of (x^3)/8

\int_{3}^{5} \frac{x ^{3}}{8} d x

17

Solution steps

\int_{3}^{5} \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_{3}^{5} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{3}^{5}

= \frac{1}{8} [\frac{x ^{4}}{4}]_{3}^{5}

Compute the boundaries:

136

= \frac{1}{8} \cdot 136

Simplify

= 17

\int_{0}^{2 π} sin (2 x) sin (x) d x

n = 0 \sum \infty (\frac{- 1}{2})^{n}

\int x^{5} - 5 x^{- 2} + 2 d x

f (x) = 4 x (x^{2} - 4 x + 5)^{9}, at x = 2

\frac{d}{d x} (9 e^{- 4 x})

What is the integral from 3 to 5 of (x^3)/8 ?
The integral from 3 to 5 of (x^3)/8 is 17