taylor 5xe^x

Lösung

taylor

5 x e^{x}

5 x + 5 x^{2} + \frac{5}{2} x^{3} + \frac{5}{6} x^{4} + \frac{5}{24} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( 5 x e ^{x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 5 x e ^{x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 5 x e ^{x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{5}{1 !} x + \frac{10}{2 !} x^{2} + \frac{15}{3 !} x^{3} + \frac{20}{4 !} x^{4} + \frac{25}{5 !} x^{5} + \dots

Fasse zusammen

= 5 x + 5 x^{2} + \frac{5}{2} x^{3} + \frac{5}{6} x^{4} + \frac{5}{24} x^{5} + \dots

\int \frac{3 x ^{2} - 5 x - 17}{( x + 3 ) ( x - 2 ) ^{2}} d x

\int (e^{x} + x^{2} - 2^{x}) d x

\int (2 x - 4) (x + 2) d x

a re a x - 4, x = 8

2 (y - 4 x^{2}) d x + x d y = 0