en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Calculus >

integral of (y-1)

Solution

\int (y - 1) d y

Solution

\frac{y ^{2}}{2} - y + C

Solution steps

\int y - 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y d y - \int 1 d y

\int y d y = \frac{y ^{2}}{2}

\int 1 d y = y

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

Add a constant to the solution

= \frac{y ^{2}}{2} - y + C

Graph

Popular Examples

(\partial)/(\partial x)(xe^{4x})

\frac{\partial}{\partial x} (x e^{4 x})

area 2x=y,x+y=4,x+y=12

a re a 2 x = y, x + y = 4, x + y = 12

integral of ((3x^2+2x-3))/(x^3-x)

\int \frac{( 3 x ^{2} + 2 x - 3 )}{x ^{3} - x} d x

derivative of t^2e^{-t}+(ln(t))^2

d er i v a t i v e t^{2} e^{- t} + (ln (t))^{2}

area y=x^2,y= 2/(x^2+1)

a re a y = x^{2}, y = \frac{2}{x ^{2} + 1}

Frequently Asked Questions (FAQ)

What is the integral of (y-1) ?
The integral of (y-1) is (y^2)/2-y+C