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

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

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integral from 0 to 1 of sin(2pix)

\int_{0}^{1} sin (2 π x) d x

0

Solution steps

\int_{0}^{1} sin (2 π x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2 π} [- cos (u)]_{0}^{2 π}

Compute the boundaries:

0

= \frac{1}{2 π} \cdot 0

Simplify

= 0

x \to 0 lim (0.9 9^{2} + 1)

\int \frac{e ^{x} + 9}{( e ^{x} - 3 ) ^{2}} d x

\frac{\partial}{\partial x} (\frac{v ^{2}}{x})

\int - ln (x) e^{- x} d x

f (x) = 4 x e^{x}, at x = 0

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