en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of t-1

Solution

\int_{0}^{1} t - 1 d t

Solution

- \frac{1}{2}

+1

Solution steps

\int_{0}^{1} t - 1 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_{0}^{1} t d t - \int_{0}^{1} 1 d t

\int_{0}^{1} t d t = \frac{1}{2}

\int_{0}^{1} 1 d t = 1

= \frac{1}{2} - 1

Simplify

\frac{1}{2} - 1 : - \frac{1}{2}

= - \frac{1}{2}

Popular Examples

integral from x^3 to x+1 of (3x+2y)

\int_{x^{3}}^{x + 1} (3 x + 2 y) d y

integral from 0 to 3 of (xsqrt(9-x^2))

\int_{0}^{3} (x 9 - x^{2}) d x

integral from pi/4 to pi of sin^2(x)

\int_{\frac{π}{4}}^{π} sin^{2} (x) d x

integral from 0 to pi of cos(2x)sin(2x)

\int_{0}^{π} cos (2 x) sin (2 x) d x

integral from 2pi to 4pi of sin(x)

\int_{2 π}^{4 π} sin (x) d x

Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of t-1 ?
The integral from 0 to 1 of t-1 is -1/2