L Lukybear Active Member Joined May 6, 2008 Messages 1,466 Gender Male HSC 2010 Feb 2, 2010 #1 Find the derivative of f(x) = cos x + x sin x. Hence evaluate the definite integral
shaon0 ... Joined Mar 26, 2008 Messages 2,029 Location Guess Gender Male HSC 2009 Feb 2, 2010 #2 Lukybear said: Find the derivative of f(x) = cos x + x sin x. Hence evaluate the definite integral Click to expand... f'(x)=-sin(x)+sin(x)+xcos(x) =xcos(x) Hence, S {x=0 to x=pi/2} xcos(x) dx = S {x=0 to x=pi/2} d/dx(cos x + x sin x) dx = {x=0 to x=pi/2} [(cos x + x sin x)] = pi/2-1
Lukybear said: Find the derivative of f(x) = cos x + x sin x. Hence evaluate the definite integral Click to expand... f'(x)=-sin(x)+sin(x)+xcos(x) =xcos(x) Hence, S {x=0 to x=pi/2} xcos(x) dx = S {x=0 to x=pi/2} d/dx(cos x + x sin x) dx = {x=0 to x=pi/2} [(cos x + x sin x)] = pi/2-1
L Lukybear Active Member Joined May 6, 2008 Messages 1,466 Gender Male HSC 2010 Feb 2, 2010 #3 How did you integrate the cosx+xsinx? O crud, nvm
L Lukybear Active Member Joined May 6, 2008 Messages 1,466 Gender Male HSC 2010 Feb 2, 2010 #4 Also Derive f(x) = sin2x -2x.cos2x, and us this to find S{pi/4,0} xsin2x dx
L life92 Member Joined May 7, 2008 Messages 81 Gender Male HSC 2009 Feb 2, 2010 #5 Derive f(x) = sin2x -2x.cos2x, and us this to find S{pi/4,0} xsin2x dx f'(x) = 2cos2x - 2 (cos2x -2xsin2x) = 4xsin2x therefore, S{pi/4,0} xsin2x dx = 1/4 [ sin2x - 2xcos2x ] {pi/4,0} = 1/4 [ (1 - 0) - (0 - 0) = 1/4
Derive f(x) = sin2x -2x.cos2x, and us this to find S{pi/4,0} xsin2x dx f'(x) = 2cos2x - 2 (cos2x -2xsin2x) = 4xsin2x therefore, S{pi/4,0} xsin2x dx = 1/4 [ sin2x - 2xcos2x ] {pi/4,0} = 1/4 [ (1 - 0) - (0 - 0) = 1/4
shaon0 ... Joined Mar 26, 2008 Messages 2,029 Location Guess Gender Male HSC 2009 Feb 2, 2010 #6 Lukybear said: Also Derive f(x) = sin2x -2x.cos2x, and us this to find S{pi/4,0} xsin2x dx Click to expand... f'(x)=2cos(2x)-2cos2x+4x.sin2x =4xsin(2x) S{pi/4,0} xsin2x dx=(1/4)S{pi/4,0}4xsin2x dx =(1/4) S{pi/4,0} d/dx(sin2x -2x.cos2x) dx =(1/4) {pi/4,0}sin2x -2x.cos2x =(1/4)
Lukybear said: Also Derive f(x) = sin2x -2x.cos2x, and us this to find S{pi/4,0} xsin2x dx Click to expand... f'(x)=2cos(2x)-2cos2x+4x.sin2x =4xsin(2x) S{pi/4,0} xsin2x dx=(1/4)S{pi/4,0}4xsin2x dx =(1/4) S{pi/4,0} d/dx(sin2x -2x.cos2x) dx =(1/4) {pi/4,0}sin2x -2x.cos2x =(1/4)