Question

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.∫₋₂¹ (1 / x⁴) dx

Accepted Answer

Identify the type of improper integral: Since the integrand is $$ \frac{1}{x^4} $$ and the interval is from $$ -2  to$$ $$ 1 $$, note that the function is undefined at $$ x = 0 $$ because division by zero is not allowed. This means the integral is improper due to a discontinuity at $$ x = 0 . $$Split the integral at the point of discontinuity to handle it properly: Write the integral as the sum of two integrals, $$ \int_{-2}^{0} \frac{1}{x^4} \, dx + \int_{0}^{1} \frac{1}{x^4} \, dx . $$Each of these integrals will be evaluated as limits approaching 0 from the left and right, respectively. Set up the limits for each integral: For the left integral, write $$ \lim_{t 	o 0^-} \int_{-2}^{t} \frac{1}{x^4} \, dx $$, and for the right integral, write $$ \lim_{s 	o 0^+} \int_{s}^{1} \frac{1}{x^4} \, dx . $$Find the antiderivative of $$ \frac{1}{x^4} $$: Recall that $$ \frac{1}{x^4} = x^{-4} . $$The antiderivative is $$ \int x^{-4} \, dx = \frac{x^{-3}}{-3} + C = -\frac{1}{3x^3} + C . $$Evaluate each integral using the antiderivative and then take the limits: Substitute the limits into the antiderivative expressions for both integrals, then analyze the behavior as $$ t 	o 0^- $$ and $$ s 	o 0^+  to $$determine if the limits exist (converge) or not (diverge). If both limits exist and are finite, sum them to find the value of the original integral.

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.
∫₋₂¹ (1 / x⁴) dx

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Improper Integrals

Behavior of Functions Near Discontinuities

Evaluating Limits for Convergence