Question

\(5 - 32\)Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

10. \(\int_{ - \infty }^0 {{2^r}} dr\)

Short answer

Integral is convergent & converges to\(\frac{1}{{\ln 2}}\).

Step-by-step solution

Definition for converging integral

Consider the integral\(\int_{\bf{a}}^{\bf{t}} {\bf{f}} {\bf{(x)dx}}\)is for the limit with the finite number\(t \ge a\). Then, the integral is written as:

\(\int_a^\infty f (x)dx = \mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)

Evaluate the integral

In order to determine if integral converges or diverges we evaluate both integrals i.e; \(\int_a^\infty f (x)dx\,\& \,\mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\) and see if both integrals are equal or not.

Given integral can be rewritten as:

\(\int_{ - \infty }^0 {{2^r}} dr = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{2^r}} dr\)

Evaluate\(\int_a^0 {{2^r}} dr\)using exponential rule i.e; \(\int {{b^x}} dx = {b^x} \times \frac{1}{{\ln b}}\)

So, we have:

\(\begin{aligned}{c}\int_a^0 {{2^r}} dr &= \frac{1}{{\ln 2}} \times \left. {{2^r}} \right|_a^0\\ &= \frac{1}{{\ln 2}} \times \left( {{2^0} - {2^a}} \right)\\ &= \frac{1}{{\ln 2}} \times \left( {1 - {2^a}} \right)\end{aligned}\)

Evaluate limit:

Now evaluate the limit as\(a \to - \infty \) in the above integral.

\(\begin{aligned}{c}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{2^r}} dr &= \mathop {\lim }\limits_{a \to - \infty } \frac{1}{{\ln 2}} \times \left( {1 - {2^a}} \right)\\ &= \frac{1}{{\ln 2}} \times \mathop {\lim }\limits_{a \to - \infty } \left( {1 - {2^a}} \right)\end{aligned}\)

The limit of a sum is the sum of the limits if they both exist and are finite, so

\(\begin{aligned}{c}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{2^r}} dr &= \frac{1}{{\ln 2}} \times \left( {\mathop {\lim }\limits_{a \to - \infty } 1 - \mathop {\lim }\limits_{a \to - \infty } {2^a}} \right)\\ &= \frac{1}{{\ln 2}} \times (1 - 0)\\ &= \frac{1}{{\ln 2}}\end{aligned}\)

As limit is finite so the integral is convergent & \(\int_{ - \infty }^0 {{2^r}} dr = \frac{1}{{\ln 2}}\).