What is the integral of (1/8 x)x^2 ?

The integral of (1/8 x)x^2 is (x^4)/(32)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1/8 x)x^2

\int (\frac{1}{8} x) x^{2} d x

\frac{x ^{4}}{32} + C

Solution steps

\int \frac{1}{8} x x^{2} 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 x x^{2} d x

Simplify

x x^{2} : x^{3}

= \frac{1}{8} \cdot \int x^{3} d x

Apply the Power Rule

: \frac{x ^{4}}{4}

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

Simplify

\frac{1}{8} \cdot \frac{x ^{4}}{4} : \frac{x ^{4}}{32}

= \frac{x ^{4}}{32}

Add a constant to the solution

= \frac{x ^{4}}{32} + C

\frac{d}{d x} (arcsin (sin (13 x)))

\int arcsec (t) d t

\frac{d}{d x} (\frac{1}{4 - 3 x})

- \frac{x}{y}

\int sin (x) 1 + cos^{2} (x) d x