Jan 28, 2009 #2 D DavidG Member May 14, 2008 205 0 16 hey mate, just to let you know, whether or not this can be achieved is based on the value of k, k must be negative for the integral to be finite, however assume k is < 0, i.e. k = -b ; b>0 Then, int (te^(kt) dt ; t = 0 to inf) = int(te^(-bt) dt ; t = 0 to inf) Employ Integration by parts int(uv') = uv - int(vu') Here let, u = t --> u' = 1 v' = e^(-bt) --> v = (-1/b)e^(-bt) Thus, int(te^(-bt) dt ; t = 0 to inf) = [ (-t/b)e^(-bt) ; t = 0 to inf ] - int( (-1/b)e^(-bt) (1) dt ; t = 0 to inf) = (0 - 0) + (1/b)int(e^(-bt) dt ; t = 0 to inf) = (1/b) [ (-1/b)e^(-bt) ; t = 0..int ] = (1/b) [ (-1/b) - 0] = -1/b^2 Hope this helps, David
hey mate, just to let you know, whether or not this can be achieved is based on the value of k, k must be negative for the integral to be finite, however assume k is < 0, i.e. k = -b ; b>0 Then, int (te^(kt) dt ; t = 0 to inf) = int(te^(-bt) dt ; t = 0 to inf) Employ Integration by parts int(uv') = uv - int(vu') Here let, u = t --> u' = 1 v' = e^(-bt) --> v = (-1/b)e^(-bt) Thus, int(te^(-bt) dt ; t = 0 to inf) = [ (-t/b)e^(-bt) ; t = 0 to inf ] - int( (-1/b)e^(-bt) (1) dt ; t = 0 to inf) = (0 - 0) + (1/b)int(e^(-bt) dt ; t = 0 to inf) = (1/b) [ (-1/b)e^(-bt) ; t = 0..int ] = (1/b) [ (-1/b) - 0] = -1/b^2 Hope this helps, David
Jan 28, 2009 #3 D DavidG Member May 14, 2008 205 0 16 hey mate, just to let you know, whether or not this can be achieved is based on the value of k, k must be negative for the integral to be finite, however assume k is < 0, i.e. k = -b ; b>0 Then, int (te^(kt) dt ; t = 0 to inf) = int(te^(-bt) dt ; t = 0 to inf) Employ Integration by parts int(uv') = uv - int(vu') Here let, u = t --> u' = 1 v' = e^(-bt) --> v = (-1/b)e^(-bt) Thus, int(te^(-bt) dt ; t = 0 to inf) = [ (-t/b)e^(-bt) ; t = 0 to inf ] - int( (-1/b)e^(-bt) (1) dt ; t = 0 to inf) = (0 - 0) + (1/b)int(e^(-bt) dt ; t = 0 to inf) = (1/b) [ (-1/b)e^(-bt) ; t = 0..int ] = (1/b) [ (-1/b) - 0] = -1/b^2 Hope this helps, David
hey mate, just to let you know, whether or not this can be achieved is based on the value of k, k must be negative for the integral to be finite, however assume k is < 0, i.e. k = -b ; b>0 Then, int (te^(kt) dt ; t = 0 to inf) = int(te^(-bt) dt ; t = 0 to inf) Employ Integration by parts int(uv') = uv - int(vu') Here let, u = t --> u' = 1 v' = e^(-bt) --> v = (-1/b)e^(-bt) Thus, int(te^(-bt) dt ; t = 0 to inf) = [ (-t/b)e^(-bt) ; t = 0 to inf ] - int( (-1/b)e^(-bt) (1) dt ; t = 0 to inf) = (0 - 0) + (1/b)int(e^(-bt) dt ; t = 0 to inf) = (1/b) [ (-1/b)e^(-bt) ; t = 0..int ] = (1/b) [ (-1/b) - 0] = -1/b^2 Hope this helps, David
Jan 28, 2009 #4 5 5aboob New member Jan 28, 2009 1 0 1 by parts: u= t ---> du = dt dv = e^(kt) dt ----> v= 1/k e^kt uv - int v du = t/k e^kt - int 1/k e^kt dt = t/k e^kt - 1/k^2 e^kt + c for the given limits the answer is infinity
by parts: u= t ---> du = dt dv = e^(kt) dt ----> v= 1/k e^kt uv - int v du = t/k e^kt - int 1/k e^kt dt = t/k e^kt - 1/k^2 e^kt + c for the given limits the answer is infinity