Three friends (a b and c) will participate in a round robin tournament in which each one plays both of the others. Suppose that p (a beats b) equals .7 (a beats c) .8 and p(b beats c )= .6 and the outcomes of the matches are independent A)what is the probability that each person wins exactly one match?

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To solve this integral, what you can do is split it up as follows: (1) ∫40(x2−4)dx=∫40(x2)dx−∫40(4)dx. to integrate  4  as an  indefinite integral, a good rule of thumb is  by  an equation below  for  polynomials; . (2) ∫xa=xa+1a+1+c , where a is any number. c denotes some constant that is crucial when doing 

The answer is    (d) 7/2======we have    we also  have an integral      notice how function f splits up at x = 0. there's a left piece and a right pieceso we'll need to do the same thing with our integral: split it at x = 0we use a property of integrals that allows us to split up and sum:     notice that for the piecewise function f, the function has the equationf(x) = x when x  ≤ 0. so for the integral  , f(x) = x so it changes into      for the piecewise function f, the function has the equationf(x) = x+1  when x  > 0. for the integral  , we have    therefore, back to our splitted integral: