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

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

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integral from 0 to 1 of sin(t)

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

- cos (1) + 1

Decimal

0.45969 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- cos (1) + 1

= - cos (1) + 1

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

\int_{0}^{4} (\frac{1}{2 x + 1}) d x

\int_{2 π}^{3 π} sin (x) d x

\int_{0}^{2} e^{- y} d y

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

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