Mathematics I20795 marksNumerical: improper integral convergenceImproper integrals
Show the integral coverages int03 fracdxx-1.
Show the integral coverages $\int_0^3 \frac{dx}{x-1}$. [5]
- Integral: $\displaystyle\int0^3 \frac{dx}{x-1}$ - Integrand: $f(x) = \dfrac{1}{x-1}$ - Limits: lower $= 0$, upper $= 3$ The denominator vanishes when $x - 1 = 0$, i.e. at $x = 1$. Since $1 \in (0,3)$, the integrand has an infinite discontinuity inside the interval. This is an improper integral ...