L\{\sqrt{{t}/{\pi}}\} = \frac{1}{s^(3/2)}
and
L\{\cos(8t)\} = \frac{s}{s^2 + 8^2}
So we are looking for ( 1 / (pi^(1/2)) \int (t-v)^(1/2) cos(8v) dv Integrating from O to t
1. L-1{(3s+2)/ (s2+2s +10)}
After completing the square I get to 3s+2 /(s+1)2 + 32 which suggests two solutions or one. They decompose the fraction into [(A)s+1 /(s+1)2 + 32 ]+ [(B) 3/(s+1)2 + 32]
I am unsure of how this decomposition works I thought that we would take A(3s) as the numerator...
Yes, not 2 equal halves but the rotation around the x=1 axis would cover the right hand side of the curve in rotation and double over so can't we ignore that piece?
Find the volume of the solid st,
1. y=cos x , y= 0 in [0,pi] ; Rotated around x=1
2. I am slightly confused, I see that the area will double around twice so I can just use the left half of the curve. I am just not sure how to do so.
1.\int2x^2+x+9/(9x+1)(x^2+9) dx
2. (A/9x+1) + [(Bx + C ) / (x^2 + 9)]
I get the worst numbers when I solve the system. The question is from an old exam and calculators are not allowed. Am I doing something wrong or is there another way to integrate this?
1. f(x)= 3x2-12x / x2-6x +8
f(x) can be made continuous at x =4 by defining f(4)=6
I know that the removable disc. is at x=2 and the non removable is at x=4. So there is an asymptote at x=4. How is it possible to define it there?