\int_{0}^{\infty} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}
Mar 19, 2007 #1 catcherintherye 48 0 [tex]\int_{0}^{\infty} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}[/tex] Last edited by a moderator: Mar 20, 2007
Mar 20, 2007 #3 HallsofIvy Science Advisor Homework Helper 41,847 969 catcherintherye said: [tex]\int_{0}^{\infty} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}[/tex] Is there a question here? My question is 'how did the [itex]\infty[/itex] in the integral become 1 in the evaluation? Also, in what sense does this have anything to do with a "Laplace transform"? could you have forgotten an [itex]e^{-xt}[/itex]? Last edited by a moderator: Mar 20, 2007
catcherintherye said: [tex]\int_{0}^{\infty} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}[/tex] Is there a question here? My question is 'how did the [itex]\infty[/itex] in the integral become 1 in the evaluation? Also, in what sense does this have anything to do with a "Laplace transform"? could you have forgotten an [itex]e^{-xt}[/itex]?