Whats the convolution of this x(t) and h(t)?

x(t)= e^(-.25*t) [u(t)-u(t-6)]

h(t)= e^-(t+1) [u(t+1)-u(t-5)]

The convolution of h(t)*x(t) is

h(t)*x(t) = ∫h(τ)x(t-τ)dτ evaluated from - ∞ to ∞

So you have to evaluate

∫( e^-(τ+1) [u(τ+1)-u(τ-5)])( e^(-.25*(t-τ)[u(t-τ)-u(t-τ-6)])dτ

which is after modifying the x(t-τ) term

∫( e^-(τ+1) [u(τ+1)-u(τ-5)])( e^(.25*(τ-t)[u(t-τ)-u(t-τ-6)])dτ

simplifying by combining the exponent terms gives

∫( e^-(0.75τ+1+t) [u(τ+1)-u(τ-5)][u(t-τ)-u(t-τ-6)])dτ

when τ < -1, convolution integral is 0

and when τ ≥ 5, the convolution integral is 0

Furthermore τ ≤ t < τ +6

t must be between -1 and 11 when t <-1, the convolution integral is 0 and when t >11, the convolution integral is also 0

So evaluate the integral below from τ = -1 to t-6 ( which means 5 < t ≤ 11)

∫( e^-(0.75τ+1+t)dτ = -0.75(e^-(0.75τ +1+t)

if t =6, the above integral is evaluated from τ = -1 to 0

x(6)*h(6) = -0.75(e^-7) +0.75e^ -(0.75(-1)+1+6) = 0.75(e^(-6.25) -e^-7)

Double check this