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

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

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integral from 0 to 1 of sin(u)

\int_{0}^{1} sin (u) d u

- cos (1) + 1

Decimal

0.45969 \dots

Solution steps

\int_{0}^{1} sin (u) d u

Use the common integral:

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

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

Compute the boundaries:

- cos (1) + 1

= - cos (1) + 1

\int_{0}^{3} 24 - 8 x d x

\int_{0}^{3} \frac{1}{5} e^{- \frac{x}{5}} d x

\int_{0}^{4} (2 t - 1)^{2} d t

\int_{- 6}^{0} (4 + 36 - x^{2}) d x

\int_{- 1}^{1} 1 - 2 x^{2} + x^{4} d x

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