91Ó°ÊÓ

An improper integral occurs when one or both limits of integration are infinite. In this case, the improper integral can be defined by replacing the infinite limits with variables. The improper integral is then evaluated by taking the limit of the integral as the variables approach infinity. For example, consider the integral: $$ \int_{1}^{\infty} \frac{1}{x^2} dx $$ To solve this, we replace the infinite limit with a variable \(b\): $$ \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} dx $$ Evaluate the integral, then take the limit: $$ \lim_{b \to \infty} \left[ -\frac{1}{x} \right]_1^b = \lim_{b \to \infty} \left( -\frac{1}{b} + \frac{1}{1} \right) = 1 $$

An improper integral occurs when the integrand has an undefined value at one or more points in the interval of integration. In this case, the improper integral can be defined by taking the limit of one or both endpoints of a continuous sub-interval as they approach the point(s) of discontinuity. For example, consider the integral: $$ \int_{0}^{1} \frac{1}{\sqrt{x}} dx $$ The function \(\frac{1}{\sqrt{x}}\) is undefined at \(x=0\). To solve this, we replace the endpoint of the discontinuous point with a variable \(a\): $$ \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} dx $$ Evaluate the integral, then take the limit: $$ \lim_{a \to 0^+} \left[ 2\sqrt{x} \right]_a^1 = \lim_{a \to 0^+} \left( 2\sqrt{1} - 2 \sqrt{a} \right) = 2 $$ The two general ways in which an improper integral may occur are: 1) having infinite limits, and 2) encountering discontinuity in the function within the interval of integration.