Integration Indefinite

Integration with Bounds

We just finished covering a large collection of integration techniques. Now we are going to spend a little bit of time putting these integrals to use in geometric problems; that is, measuring areas and volumes. To do so, we will mainly be dealing with definite integrals.

Basic Areas:

Integrals measure area between curves! How to find the area between two? Top-Bottom. What about if they are functions of $y$? Right-Left (both cases, bigger values vs smaller values).

What to do when curves swap places? Need to break up the integral into two pieces.

Example: Area between $y=x$ and $y=x^2-1$ from $x=-1$ to $x=3$.

Example: Area of the region bounded by $x^2-1$ and $1-x^2$

Areas going off to Infinity

Oftentimes you actually want to measure the area of a region which is unbounded on the $x$ axis, so an integral of the form $$\int_a^\infty f(x)dx$$ or similarly $\int_{-\infty}^a$ or $\int_{-\infty}^\infty$. These integrals measure things like the *total contribution* of some process into the indefinite future, but since $\infty$ is not a number we can’t evaluate these integrals in the usual way of plugging in the bounds.
Instead, as is common when dealing with infinity, we need to take a limit:

Definition The improper integral $\int_a^\infty f(x)dx$ is calculated as $$\lim_{t\to\infty}\int_a^t f(x)dx$$ This is called *convergent* if the limit exists, and *divergent* otherwise. The same holds for an integral with lower bound $-\infty$, we take a limit as $t\to-\infty$. To evaluate an integral with limits $-\infty$ and $\infty$, we break it at some arbitrary real number $a$ and define $$\int_{-\infty}^\infty f(x)dx=\lim_{s\to-\infty}\int_{s}^af(x)dx+\lim_{t\to\infty}\int_a^t f(x)dx$$

Examples: $$\int_1^\infty \frac{1}{x}dx$$ $$\int_1^\infty\frac{1}{x^2}dx$$

$$\int_{-\infty}^0 xe^xdx$$ $$\int_{-\infty}^\infty \frac{1}{1+x^2}dx$$

Integrating functions with a discontinuity.

If $f(x)$ is not defined at $a$, or is discontinuous at $a$, then the integral $\int_a^bf(x)dx$ is evaluated as a limit: $$\int_a^bf(x)dx=\lim_{t\to a+}\int_t^bf(x)dx$$

Example: $$\int_0^5\ln(x)dx$$

If $c\in [a,b]$ is a point where $f$ is discontinuous, then to evaluate $\int_a^b f(x)dx$ we need to break it into two integrals each of which has $c$ as an endpoint, and then evaluate each of these separately. $$\int_a^bf(x)dx=\lim_{s\to c^-}\int_a^t f(x)dx+\int_{t\to c^+}\int_t^b f(x)dx$$

Example: Evaluate $\int_0^3\frac{1}{x-1}dx$ Correct answer: integral is divergent! Function is discontinuous at $x=1$, and neither integral converges!

Comparison

Sometimes all that matters is if a particular integral converges or diverges. In these situations we can use comparison of our integral with an easier integral to determine its behavior.

If $0\leq f\leq g$ and the integral of $g$ converges, then the integral of $f$ converges.

If $0\leq f\leq g$ and the integral of $f$ diverges, then the integral of $g$ diverges.

Example: