Final answer:
The improper integral ∫∞−1 e2tdt converges, with its value being ½/e2.
Step-by-step explanation:
The question asks to compute the value of the improper integral ∫∞−1 e2tdt. To do this, we need to find the antiderivative of the function e2t and evaluate it at the limits of integration −∞ and −1.
First, we find the antiderivative:
∫ e2tdt = ½e2t + C,
where C is the constant of integration. Now, evaluating this at −1 and taking the limit as t approaches −∞:
∫∞−1 e2tdt = limt → −∞ ½e2t - ½e2(−1)
Since e2t approaches zero as t approaches negative infinity, the limit of the first term is zero, and we are left with:
½e−2 - 0 = ½/e2
Therefore, the value of the integral is ½/e2, which is a finite number, meaning the integral converges.