What is the integral from-1 to 0 of sin(x) ?

The integral from-1 to 0 of sin(x) is -1+cos(1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 0 of sin(x)

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

- 1 + cos (1)

Decimal

- 0.45969 \dots

Solution steps

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

Use the common integral:

\int sin (x) d x = - cos (x)

= [- cos (x)]_{- 1}^{0}

Compute the boundaries:

- 1 + cos (1)

= - 1 + cos (1)

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

\int_{x}^{8} \frac{1}{1 + t ^{2}} d t

\int_{0}^{10} x e^{- 2 x} d x

\int_{0}^{2} \frac{x}{x - 2} d x

\int_{1}^{e^{3}} \frac{4 ln ( x )}{x} d x

What is the integral from-1 to 0 of sin(x) ?
The integral from-1 to 0 of sin(x) is -1+cos(1)