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

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

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integral from 0 to pi of 2pix(sin(x))

\int_{0}^{π} 2 π x (sin (x)) d x

2 π^{2}

Decimal Notation

19.73920 \dots

Solution steps

\int_{0}^{π} 2 π x sin (x) d x

Take the constant out:

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

= 2 π \cdot \int_{0}^{π} x sin (x) d x

Apply Integration By Parts

: [- x cos (x) - \int - cos (x) d x]_{0}^{π}

= 2 π [- x cos (x) - \int - cos (x) d x]_{0}^{π}

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

= 2 π [- x cos (x) - (- sin (x))]_{0}^{π}

Simplify

= 2 π [- x cos (x) + sin (x)]_{0}^{π}

Compute the boundaries:

π

= 2 ππ

Simplify

= 2 π^{2}

\int cos (2 + t) d t

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 5

\int_{0}^{\infty} e^{- x^{2}} d x

\int \frac{- 5}{( x ^{2} + 5 ) ^{\frac{3}{2}}} d x

\frac{d}{d t} (\frac{6}{5} - t^{2})

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