Sie erhalten ∫ x3 = 1/(31) * x31 = = 1/2 x2 = 1/2 * 1/x² = 1/(2x²), um noch einige andere Schreibweisen zu zeigen, sowie in der etwas umständlicheren Schreibweise 1/2 * 1/x^2 Fazit Gebrochen rationale Funktionen der Art 1/x^m lassen sich recht einfach integrieren, wenn man diese in eine Funktion mit negativer Potenz umwandelt und dann die bekannte Integralregel anwendet9 x 2 = 10 x 2 = 1 x 3 = 2 x 3 = 3 x 3 = 4 x 3 = 5 x 3 = 6 x 3 = 7 x 3 = 8 x 3 = 1 x 3 = 3 10 x 2 = 9 x 2 = 18 4 x 3 = 12 3 x 3 = 9 2 x 3 = 6 7 x 3 = 21Example 1 X and Y are jointly continuous with joint pdf f(x,y) = ˆ cx2 xy 3 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 0, otherwise (a) Find c (b) Find P(X Y ≥ 1) (c) Find marginal pdf's of X and of Y (d) Are X and Y independent (justify!) (e) Find E(eX cosY) (f) Find cov(X,Y) We start (as always!) by drawing the support set (See below, left) 2 1 2 1 1 x y=1−x y x y support set
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Rd Sharma Solutions For Class 8 Chapter 9 Linear Equation In One Variable Exercise 9 1 Access Free Pdf