= 23 e5 e2 = 16 e7 cos Solution Recall that the delta function when placed in an integral actually only evaluates the integrand at a specific pt. In this case we have f(x) (x+1)dx = f(-1) So that complex mess of an integrand because very simple. 2-3x ex+6 e-2x Lne-2x cos [/3+/6](x+1) dx simplifies to just evaluating the integrand at x=-1 Likewise for all the other problems we perform the same procedure for b at x=4 for c at t=3 So on and so forth.