What is the integral from 0 to 4 of 1/96 xy ?

The integral from 0 to 4 of 1/96 xy is y/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of 1/96 xy

\int_{0}^{4} \frac{1}{96} x y d x

\frac{y}{12}

Solution steps

\int_{0}^{4} \frac{1}{96} x y d x

Take the constant out:

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

= \frac{1}{96} y \cdot \int_{0}^{4} x d x

Simplify

\frac{1}{96} y : \frac{y}{96}

= \frac{y}{96} \cdot \int_{0}^{4} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{4}

= \frac{y}{96} [\frac{x ^{2}}{2}]_{0}^{4}

Compute the boundaries:

8

= \frac{y}{96} \cdot 8

Simplify

= \frac{y}{12}

\int_{- 1}^{2} x^{2} - 2 x + 2 d x

\int_{0}^{0.4} 2 x - 5 x^{2} d x

\int_{a}^{b} sin^{2} (x) d x

\int_{0}^{4} x^{2} 1 + 2 x d x

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

What is the integral from 0 to 4 of 1/96 xy ?
The integral from 0 to 4 of 1/96 xy is y/(12)