What is the integral from y to 1 of 3y ?

The integral from y to 1 of 3y is (3(1-y^2))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from y to 1 of 3y

\int_{y}^{1} 3 y d y

\frac{3 ( 1 - y ^{2} )}{2}

Solution steps

\int_{y}^{1} 3 y d y

Take the constant out:

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

= 3 \cdot \int_{y}^{1} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{y}^{1}

= 3 [\frac{y ^{2}}{2}]_{y}^{1}

Compute the boundaries:

\frac{1}{2} - \frac{y ^{2}}{2}

= 3 (\frac{1}{2} - \frac{y ^{2}}{2})

Simplify

= \frac{3 ( 1 - y ^{2} )}{2}

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

\int_{- 1}^{1} \frac{6}{x ^{2} + 1} d x

\int_{2}^{4} 2^{5 x} d x

\int_{\frac{π}{3}}^{π} cos (x) d x

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

What is the integral from y to 1 of 3y ?
The integral from y to 1 of 3y is (3(1-y^2))/2