fläche 1/(x+1),((x-1))/2 ,0,4

Lösung

fläche

\frac{1}{x + 1}, \frac{( x - 1 )}{2}, 0, 4

ln (1 + 3) + \frac{ln ( \frac{25 ( 2 - 3 )}{2} ) + 2 + 4 3}{4} - \frac{3}{2} ln (\frac{5 ( 3 - 1 )}{2})

Dezimale

2.63271 \dots

Schritte zur Lösung

Wenden Sie die Flächenformel an:

\int_{0}^{4} \frac{1}{x + 1} - \frac{1}{2} (x - 1) d x

Löse

\int_{0}^{4} \frac{1}{x + 1} - \frac{1}{2} (x - 1) d x : ln (1 + 3) + \frac{ln ( \frac{25 ( 2 - 3 )}{2} ) + 2 + 4 3}{4} - \frac{3}{2} ln (\frac{5 ( 3 - 1 )}{2})

Die Fläche ist:

= ln (1 + 3) + \frac{ln ( \frac{25 ( 2 - 3 )}{2} ) + 2 + 4 3}{4} - \frac{3}{2} ln (\frac{5 ( 3 - 1 )}{2})

a re a x^{2}, 3 x

\int 2 y^{2} cos (9 y) d y

s l o p e y^{5} + 3 x^{2} y^{2} + 5 x^{4} = 49, (- 1, 2)

y^{^{'}} = (t y^{2} - 1) y, y (1) = \frac{1}{2}

\int 1 - e^{3 x} d x