What is the integral from 0 to 2 of sin(pix) ?

The integral from 0 to 2 of sin(pix) is 0

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integral from 0 to 2 of sin(pix)

\int_{0}^{2} sin (π x) d x

0

Solution steps

\int_{0}^{2} sin (π x) d x

Apply u-substitution

: \int_{0}^{π 2} sin (u) \frac{1}{π} d u

= \int_{0}^{π 2} sin (u) \frac{1}{π} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{π} \cdot \int_{0}^{π 2} sin (u) d u

Use the common integral:

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

= \frac{1}{π} [- cos (u)]_{0}^{π 2}

Compute the boundaries:

0

= \frac{1}{π} \cdot 0

Simplify

= 0

\int \frac{π}{x ^{2}} d x

\int \frac{2}{x ^{2} + x + 1} d x

\int 9 - t^{2} d t

x + 2, x^{5} + 2

\frac{d}{d x} (- 2 sec^{2} (x))

What is the integral from 0 to 2 of sin(pix) ?
The integral from 0 to 2 of sin(pix) is 0