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

inverse oflaplace 3/((s+2)^2+9)inverse laplace

\frac{3}{( s + 2 ) ^{2} + 9}

integral from 0 to pi of sin^5(x/2)

\int_{0}^{π} sin^{5} (\frac{x}{2}) d x

limit as t approaches pi of sin(t/6)

t \to π lim (sin (\frac{t}{6}))

(dy)/(dx)=2ysqrt(16-x^2),y(0)=1

\frac{d y}{d x} = 2 y 16 - x^{2}, y (0) = 1

derivative of ln(16-x^2)

\frac{d}{d x} (ln (16 - x^{2}))

Frequently Asked Questions (FAQ)

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