What is the integral from 0 to 1-y of 1 ?

The integral from 0 to 1-y of 1 is -y+1

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-y of 1

\int_{0}^{1 - y} 1 d z

- y + 1

Solution steps

\int_{0}^{1 - y} 1 d z

Integral of a constant:

\int a d x = a x

= [1 \cdot z]_{0}^{1 - y}

Simplify

= [z]_{0}^{1 - y}

Compute the boundaries:

- y + 1

= - y + 1

\int_{0}^{1} x (1 - x)^{9} d x

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

\int_{1}^{2} 1 - 4 e^{- 3 x} d x

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

\int_{- 1}^{32} \frac{1}{x ^{\frac{1}{5}}} d x

What is the integral from 0 to 1-y of 1 ?
The integral from 0 to 1-y of 1 is -y+1