1. Calculate the integral (dydx) where y = 2 , y = 4-2x and x= 0 , x = 1?

Question

1. Calculate the integral (dydx) where y = 2 , y = 4-2x and x= 0 , x = 1

2. Calculate the integral (4-2u /v^2)dvdu where v = -2 , v = 4-2u and u= 0 , u = 3

Answer 1

1. This is a problem in double integration. The inner integral is integ(dy) from 2 to 4 - 2x; the antiderivative is y evaluated between the given limits, so (4 - 2x) - 2 or 2 - 2x. Now your problem is to find integ(2 - 2x)dx from 0 to 1. the antiderivative is 2x - x^2; the evaluation is (2 - 1^2) - (0 - 0) = 1.

2 Here, integ((4 - 2u)/v^2)dv = (4 - 2u/(-v) and the evaluation is (4 - 2u)/(-(4 - 2u)) - (4 - 2u)/2 = -1 - (2 - u) = -3 + u. Now
integ(-3 + u) = -3u + (u^2)/2 froim 0 to 3. The evaluation is
-9 + 9/2 = -9/2.

Answer 2

Rearranging the 1st equation we see that y = a million - x. exchange this into the 2d equation: 2x - y = 2 2x - (a million - x) = 2 2x - a million + x = 2 3x - a million = 2 3x = 3 x = a million Now all of us understand what x is, we can use the 1st equation to be sure what y is: x + y = a million a million + y = a million y = 0 And so the respond is x = a million, y = 0

Answer 3

Why?