What is the integral from 0 to 1 of x(1-x)sin(npix) ?

The integral from 0 to 1 of x(1-x)sin(npix) is (-2(-1)^n+2)/(pi^3n^3)

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integral from 0 to 1 of x(1-x)sin(npix)

\int_{0}^{1} x (1 - x) sin (nπ x) d x

\frac{- 2 ( - 1 ) ^{n} + 2}{π ^{3} n ^{3}}

Solution steps

\int_{0}^{1} x (1 - x) sin (nπ x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

- \frac{2 ( - 1 ) ^{n}}{π ^{3} n ^{3}} + \frac{2}{π ^{3} n ^{3}}

= - \frac{2 ( - 1 ) ^{n}}{π ^{3} n ^{3}} + \frac{2}{π ^{3} n ^{3}}

Simplify

= \frac{- 2 ( - 1 ) ^{n} + 2}{π ^{3} n ^{3}}

x^{- 7}

\int_{1}^{\infty} 3 x e^{- x} d x

\int tanh^{2} (3 x) d x

n = 0 \sum \infty n \cdot a^{n}

\frac{d}{d z} (e^{x z})

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