Type 2 - Improper Integrals with Discontinuous Integrands

An improper integral of type 2 is an integral whose integrand has a discontinuity in the interval
of integration $[a,b]$. This type of integral
may look normal, but it cannot be evaluated using FTC II, which requires a continuous
integrand on $[a,b]$.

Warning: Now that we have
introduced discontinuous integrands, you will need to check
every integrand you work with for any discontinuities on the
interval of integration.

Examples: $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$
and $\displaystyle\int_{-1}^1 \frac{dx}{x^2}$ are of type 2, since
$\displaystyle\lim_{x\to0}\frac{1}{\sqrt x}$ and
$\displaystyle\lim_{x\to0}\frac{1}{x^2}$ do not exist, and $0$ is
contained in the intervals $[0,1]$ and $[-1,1]$, respectively.

We evaluate integrals with discontinuous integrands by taking a
limit; the function is continuous as $x$ approaches the
discontinuity, so FTC II will
work.

When the discontinuity is at an endpoint
of the interval of integration $[a,b]$, we take the
limit as $t$ approaces $a$ or $b$ from inside $[a,b]$.

If the discontinuity is in the middle of
the interval of integration, we need to break
the integral at the point of discontinuity
into the sum of two integrals and take limits on both
integrals. In this case, we have $t$ approaching the
discontinuity from inside the interval of integration of each
integral.

As with infinite interval integrals, the improper integral converges if the corresponding
limit exists, and
diverges if it
doesn't. When we have to break an integral at the point of
discontinuity, the original integral converges only if both pieces
converge.

The following video explains improper integrals with discontinuous
integrands (type 2), and works a number of examples.