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

The integral from 0 to 1 of sin(pix) is 2/pi

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integral from 0 to 1 of sin(pix)

\int_{0}^{1} sin (π x) d x

\frac{2}{π}

Decimal Notation

0.63661 \dots

Solution steps

\int_{0}^{1} sin (π x) d x

Apply u-substitution

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

= \int_{0}^{π} 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} sin (u) d u

Use the common integral:

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

= \frac{1}{π} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{π} \cdot 2

Simplify

= \frac{2}{π}

\frac{\partial}{\partial y} (x e^{x + 2 y})

y^{^{'}} = 3 x^{2} y + 2 x y

\int_{1}^{100} \frac{9}{x ^{3}} d x

f (x) = 3 x^{2} + 32 x + 64

3 x^{3}, [0, 6]

What is the integral from 0 to 1 of sin(pix) ?
The integral from 0 to 1 of sin(pix) is 2/pi