What is the integral from-pi to 2 of sin(t) ?

The integral from-pi to 2 of sin(t) is -cos(2)-1

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integral from-pi to 2 of sin(t)

\int_{- π}^{2} sin (t) d t

- cos (2) - 1

Decimal

- 0.58385 \dots

Solution steps

\int_{- π}^{2} sin (t) d t

Use the common integral:

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

= [- cos (t)]_{- π}^{2}

Compute the boundaries:

- cos (2) - 1

= - cos (2) - 1

d er i v a t i v e f (x) = lo g_{5} (x)

2 y \frac{d y}{d x} = x (y^{2} + 1), y (2) = 0

\int \frac{10 x ^{2} - 9 x + 24}{x ^{3} + 4 x} d x

\frac{d y}{d x} = sin (4 x)

\frac{d}{d x} (e^{x + 4})

What is the integral from-pi to 2 of sin(t) ?
The integral from-pi to 2 of sin(t) is -cos(2)-1