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

Lösung

\int_{0}^{1} x (1 - x) sin (nπ x) d x

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

Schritte zur Lösung

\int_{0}^{1} x (1 - x) sin (nπ x) d x

Wende die partielle Integration an

= [- \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}

Vereinfache

= [- \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}

Berechne die Grenzen:

- \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}}

Vereinfache

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

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

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

\frac{\partial}{\partial x} (- 2 (y + 1) x^{- 3})

2 sin (x) y^{^{'}} + y cos (x) = y^{2} cos (x)

\frac{d}{d x} (\frac{4}{3} tan (x))