Decide whether the integral converges. If the integral converges, compute its value. $\int _ { - \infty } ^ { - 8 } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$
A) $- \frac { 1 } { 8 }$
B) $\frac { 1 } { 64 }$
C) $- \frac { 1 } { 64 }$
D) $\frac { 1 } { 8 }$
E) diverges

Question

Decide whether the integral converges. If the integral converges, compute its value.   $\int _ { - \infty } ^ { - 8 } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$  
A) $- \frac { 1 } { 8 }$
B) $\frac { 1 } { 64 }$
C) $- \frac { 1 } { 64 }$
D) $\frac { 1 } { 8 }$
E) diverges

Quizplus · Accepted Answer

To determine if the integral converges and to compute its value, we evaluate the improper integral $\int _ { - \infty } ^ { - 8 } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$. We can rewrite the integral with a limit: $\lim_{a \to -\infty} \int_a^{-8} \frac{1}{x^2} dx$. The antiderivative of $\frac{1}{x^2}$ is $-\frac{1}{x}$, so we have:$\lim_{a \to -\infty} \left[ -\frac{1}{x} \right]_a^{-8} = \lim_{a \to -\infty} \left( -\frac{1}{-8} - (-\frac{1}{a}) \right) = \frac{1}{8} - 0 = \frac{1}{8}$.Therefore, the integral converges to $\frac{1}{8}$.

Decide Whether the Integral Converges ∫−∞−81x2 dx\int _ { - \infty } ^ { - 8 } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x∫−∞−8​x21​ dx

Decide Whether the Integral Converges $\int _ { - \infty } ^ { - 8 } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$