What is the integral from 0 to 1 of x^3sin(2pix) ?

The integral from 0 to 1 of x^3sin(2pix) is 3/(4pi^3)-1/(2pi)

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integral from 0 to 1 of x^3sin(2pix)

\int_{0}^{1} x^{3} sin (2 π x) d x

\frac{3}{4 π ^{3}} - \frac{1}{2 π}

Decimal Notation

- 0.13496 \dots

Solution steps

\int_{0}^{1} x^{3} sin (2 π x) d x

Apply Integration By Parts

: [- \frac{1}{2 π} x^{3} cos (2 π x) - \int - \frac{3}{2 π} x^{2} cos (2 π x) d x]_{0}^{1}

= [- \frac{1}{2 π} x^{3} cos (2 π x) - \int - \frac{3}{2 π} x^{2} cos (2 π x) d x]_{0}^{1}

\int - \frac{3}{2 π} x^{2} cos (2 π x) d x = - \frac{3}{2 π} (\frac{1}{2 π} x^{2} sin (2 π x) - \frac{1}{π} (- \frac{1}{2 π} x cos (2 π x) + \frac{1}{4 π ^{2}} sin (2 π x)))

= [- \frac{1}{2 π} x^{3} cos (2 π x) - (- \frac{3}{2 π} (\frac{1}{2 π} x^{2} sin (2 π x) - \frac{1}{π} (- \frac{1}{2 π} x cos (2 π x) + \frac{1}{4 π ^{2}} sin (2 π x))))]_{0}^{1}

Simplify

= [- \frac{1}{2 π} x^{3} cos (2 π x) + \frac{3}{2 π} (\frac{1}{2 π} x^{2} sin (2 π x) - \frac{1}{π} (- \frac{1}{2 π} x cos (2 π x) + \frac{1}{4 π ^{2}} sin (2 π x)))]_{0}^{1}

Compute the boundaries:

\frac{3}{4 π ^{3}} - \frac{1}{2 π}

= \frac{3}{4 π ^{3}} - \frac{1}{2 π}

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\int_{1}^{3} (x y^{2} + x^{2} + y + 4) d x

\int_{a}^{b} (\frac{2}{x} + 1) d x

What is the integral from 0 to 1 of x^3sin(2pix) ?
The integral from 0 to 1 of x^3sin(2pix) is 3/(4pi^3)-1/(2pi)