C cmnalo Junior Member Joined Nov 5, 2006 Messages 61 Dec 11, 2006 #1 ∫(2x + 1) square root of (x^2 +x) dx [0,1] I know I should probably use the substitution method but I'm not sure whether to make u = 2x + 1 or u= x^2 +x? Answer: 4 square root of 2 / 3
∫(2x + 1) square root of (x^2 +x) dx [0,1] I know I should probably use the substitution method but I'm not sure whether to make u = 2x + 1 or u= x^2 +x? Answer: 4 square root of 2 / 3
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Dec 11, 2006 #2 Taking u = x^2 + x, then du = 2x + 1 dx . It should be clear that this is advantageous.
C cmnalo Junior Member Joined Nov 5, 2006 Messages 61 Dec 11, 2006 #3 so.. ∫ (squareroot of u) (du) [0,1] ∫u^1/2 du 2/3u^3/2 2/3(t+1)^3/2 [0,1] [2/3(1 +1)^3/2] - [2/3(0 +1)^3/2] Is this looking correct?
so.. ∫ (squareroot of u) (du) [0,1] ∫u^1/2 du 2/3u^3/2 2/3(t+1)^3/2 [0,1] [2/3(1 +1)^3/2] - [2/3(0 +1)^3/2] Is this looking correct?
C cmnalo Junior Member Joined Nov 5, 2006 Messages 61 Dec 12, 2006 #4 I'm still having trouble with this problem can anyone assist. Thanks
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Dec 12, 2006 #5 All you have to do is find \(\displaystyle \L \int\limits_0^2 {\sqrt u du} .\)
