What is the integral from 1 to t of-t+2 ?

The integral from 1 to t of-t+2 is -(t^2-1)/2+2t-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to t of-t+2

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

- \frac{t ^{2} - 1}{2} + 2 t - 2

Solution steps

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

Apply the Sum Rule:

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

= - \int_{1}^{t} t d t + \int_{1}^{t} 2 d t

\int_{1}^{t} t d t = \frac{t ^{2} - 1}{2}

\int_{1}^{t} 2 d t = 2 t - 2

= - \frac{t ^{2} - 1}{2} + 2 t - 2

12 t e^{10 t}

e^{3 t} sin (3 t)

\frac{d y}{d x} - 2 = - y

\frac{d}{d y} (y cos (y))

f (x) = \frac{x ^{2}}{x - 6}

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