Q94·Section II (Mathematics)·94 of 190The convolution theorem for Laplace transformsIf L{f(t)}=F(s)L\{f(t)\}=F(s)L{f(t)}=F(s) and L{g(t)}=G(s)L\{g(t)\}=G(s)L{g(t)}=G(s), then L−1{F(s)G(s)}L^{-1}\{F(s)G(s)\}L−1{F(s)G(s)} is: OptionsA∫0tf(x)g(t−x)dx\int_0^tf(x)g(t-x)dx∫0tf(x)g(t−x)dxB∫0tf(x)g(x)dx\int_0^tf(x)g(x)dx∫0tf(x)g(x)dxC∫0tf(x)g(t+x)dx\int_0^tf(x)g(t+x)dx∫0tf(x)g(t+x)dxD∫0tf(x−t)g(x−t)dx\int_0^tf(x-t)g(x-t)dx∫0tf(x−t)g(x−t)dxPick an option to check your answer.Show answerShow solution Report a problem with this question← Q93All questionsQ95 →Keyboard: A–D answer · S solution · ←/→ previous/next