Q. 19 - Calculus

Question

Some discontinuous functions are not integrable. For example, consider the function$f\left(x\right)=\frac{1}{x^2}$ on $\left[-2,2\right]$.

1.  Sketch a graph of $f$ on $\left[-2,2\right]$. What kind of discontinuity does $f$ have, and where?
2. Why might it be reasonable to think that $f$ is not integrable on $[-2,2]$ because of this discontinuity?
3. What do you think happens to a Riemann sum approximation for the area between $f$ and the $x$-axis on $\left[-2,2\right]$ as $n\to \infty$.

Step-by-Step Solution

Answer

Part $\left(a\right)$: The graph for the function $f\left(x\right)=\frac{1}{x^2}$ is,

Part $\left(b\right)$: The integration of a function is nothing but the area covered by the graph and the horizontal axis. Since, the area itself is not defined or infinite hence it is not integrable.

Part $\left(c\right)$: The Riemann sum will become the limits of integration which is expressed as an equation,

                             ${\int }_{-2}^{2}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{*}\right)\Delta x$

Part a Step 1 . Given information

$f\left(x\right)=\frac{1}{x^2}$ on $\left[-2,2\right]$.

Part a Step 2 . The objective is to sketch the graph of the function and then to determine the type of discontinuity the graph has.

From the above sketch it is clear that the graph has infinite discontinuity at $x=0$.Therefore, the graph has infinite discontinuity.

Part b Step 1 . The objective is to tell the reason to think that the function is not integrable on the interval - 2 , 2 .

The integration of a function is nothing but the area covered by the graph and the horizontal axis. Since, the area itself is not defined or infinite hence it is not integrable.

Therefore, because of the infinite area the function is not integrable.

Part c Step 1 . Riemann sum approximation for the area between f and the x-axis on [−2, 2]

As $n\to \infty$, the finite sum $\left(\sum _{}^{}\right)$of $n$ things becomes an integral $\left(\int \right)$that accumulates everything from $x=-2$ to $x=2$. The discrete list of values $f\left(x_k^{*}\right)$becomes the continuous function $f\left(x\right)$, and the small change $∆x$ becomes an “infinitesimal” change $dx$ which is expressed as an equation,

${\int }_{-2}^{2}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{*}\right)\Delta x$