The image asks, "What is wrong with the following calculation?"

The integral given is: $\int_{-1}^{3} \frac{1}{x^2} \, dx$ The steps shown evaluate it as follows: $= \left[ \frac{x^{-1}}{-1} \right]_{-1}^{3} = -\frac{1}{3} - (-1) = \frac{4}{3}$

Error in the Calculation: The calculation has a fundamental issue with the limits of integration. The integrand, $\frac{1}{x^2}$, has a discontinuity at $x = 0$, which lies within the integration interval $[-1, 3]$. When an integrand is discontinuous within the interval of integration, the integral may diverge (be undefined) rather than produce a finite result.

In this case:

• As $x$ approaches 0 from either side, $\frac{1}{x^2}$ grows unbounded.
• Thus, the integral $\int_{-1}^{3} \frac{1}{x^2} \, dx$ diverges due to the discontinuity at $x = 0$.

Therefore, this calculation is invalid because the integral does not converge over the given interval.

Would you like more details on improper integrals and dealing with discontinuities?

Related Questions:

1. What is an improper integral, and how is it evaluated?
2. Why does a discontinuity within an interval affect the convergence of an integral?
3. How do we handle integrals with singularities or points where the function is undefined?
4. What are some techniques to determine if an integral diverges?
5. Can you provide examples of other functions with similar discontinuities?

Tip:

Always check the integrand for points of discontinuity within the interval of integration before evaluating.