So first, we must solve for the antiderivative of the integrand. We will do this using u-substitution. #inte^4tdt# #u=4t# and #du=4dt# which can be rewritten as #1/4du=dt# so that we can take out the constant and perform the operation of #1/4inte^udu#. We know that the antiderivative of #e^u# is just #e^u# which leaves us with #1/4e^4t#.
The next step is to evaluate the antiderivative at the given bounds. So #1/4e^4t# evaluated at #t=-1# is #1/4e^-4#. The tricky part is to evaluate #1/4e^4t# at #=-oo#. When we plug it in, it becomes #1/4e^-oo#. That negative exponent can be brought down to the denominator of the fraction we have and it looks like #1/(4e^oo)#
The denominator becomes #oo# because an exponential that is infinitely multiplied by itself tends toward infinity and that infinity multiplied by 4 is another version of infinity so the whole fraction now looks like #1/oo# and since the denominator is infinitely large, that makes the whole value infinitely small, so we assume it to be #0#.
Now, we subtract the two values we calculated, which looks like #1/4e^-4 - 0# which just equals #1/4e^-4# and the negative exponent brings the exponential to the denominator, giving us an answer #1/(4e^4)#
