`E=a int_0^(I_0)e^(-Tx)dx`

Let `u = -Tx` then `du = -T\ dx`. Our integral is now:

`E=aint_0^(I_0)e^(-tx)dx`

`=-a/Tint_(x=0)^(x=I_0)e^udu`

`=-a/T[e^u]_(x=0)^(x=I_0)`

`=-a/T[e^(-Tx)]_0^(I_0)`

`=-a/T(e^(-TI_0)-e^0)`

`=a/T(1-e^(-TI_0))`

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