How do you show whether the improper integral #int (1/3x-6) dx# converges or diverges from negative infinity to zero?

Question

Emily Ammerman · Accepted Answer

Evaluating the integral shows that it diverges

#int_(-oo)^0(1/3x-6)dx=lim_(t->-oo)int_t^0(1/3x-6)dx# #=lim_(t->-oo)[x^2/6-6x]_t^0# #=lim_(t->-oo)(0-t^2/6+6t)# #=-oo !inRR# Therefore this integral diverges

Emily Ammerman · Answer

undefined To determine the convergence or divergence of the improper integral $\int_{-\infty}^0 \frac{1}{3x - 6} \, dx$, we need to analyze its behavior as $x$ approaches negative infinity.

First, rewrite the integral as:
$$\int_{-\infty}^0 \frac{1}{3x - 6} \, dx = \lim_{a \to -\infty} \int_a^0 \frac{1}{3x - 6} \, dx$$

Now, find the antiderivative of $\frac{1}{3x - 6}$:
$$F(x) = \frac{1}{3} \ln|3x - 6|$$

Apply the Fundamental Theorem of Calculus:
$$\lim_{a \to -\infty} F(0) - F(a)$$

$$= \lim_{a \to -\infty} \left(\frac{1}{3} \ln|3(0) - 6|\right) - \left(\frac{1}{3} \ln|3a - 6|\right)$$

$$= \lim_{a \to -\infty} \frac{1}{3} (\ln|-6| - \ln|3a - 6|)$$

$$= \lim_{a \to -\infty} \frac{1}{3} \ln\left(\frac{-6}{3a - 6}\right)$$

$$= \lim_{a \to -\infty} \ln\left(\left(\frac{-2}{a - 2}\right)^{\frac{1}{3}}\right)$$

Since the exponent approaches zero as $a$ approaches negative infinity, the expression inside the natural logarithm approaches 1.

Thus, the integral converges.

Evelyn Agard · Answer

undefined To determine whether the improper integral $ \int_{-\infty}^0 \frac{1}{3x - 6} \, dx $ converges or diverges, we first need to analyze its behavior as $ x $ approaches negative infinity and as it approaches zero.

As $ x $ approaches negative infinity, the function $ \frac{1}{3x - 6} $ approaches zero. Therefore, we need to check the behavior of the function near the singularity at $ x = 2 $.

Near $ x = 2 $, the function $ \frac{1}{3x - 6} $ behaves like $ \frac{1}{3(x-2)} $. As $ x $ approaches 2 from the left (i.e., as $ x $ approaches 2 from negative infinity), the function approaches negative infinity.

Therefore, the integral is improper at $ x = 2 $, and we need to check the convergence around this point. Since the function approaches negative infinity near $ x = 2 $, the integral diverges.

Hence, the improper integral $ \int_{-\infty}^0 \frac{1}{3x - 6} \, dx $ diverges from negative infinity to zero.