Newton Leibniz formula

Question

$$\int_{-3}^5 f(x)\,dx$$
for $$ f(x) =\begin{cases} 1/x^2, & \text{if }x \neq 0 \\ -10^{17}, & \text{if }x=0 \end{cases} $$

I tried with Newton Leibniz formula, is this correct ?

$\int_{-3}^0 f(x)dx$ + $\int_{0}^5 f(x)dx$ =

$1/x^2 |_{-3}^{0} $ $ + $ $1/x^2 |_0^5$=

$3/(-3)^2+10^{17}+(-10^{17})-3/5^2)$= $16/75$

I know I made a mistake, but I dont know what, could someone please correct me and help me.

Answer 1

The $-10^{17}$ is a red herring: the value of a function at a single point has no effect on the integral. On the other hand, this does signal that something funny is likely to be happening as $x \to 0$, and indeed it does: the integrand goes to $\infty$ there. Since the integrand is unbounded, this is not an ordinary Riemann integral, but rather an improper integral. For that, you want to take a limit:

$$ \int_{-3}^5 f(x)\; dx = \lim_{a \to 0-} \int_{-3}^a f(x)\; dx + \lim_{b \to 0+} \int_b^5 f(x)\; dx $$ However, (if you use a correct antiderivative) you'll find that both of these limits are $+\infty$. So the conclusion is that the improper integral does not exist (or is $+\infty$, depending on your point of view).

Answer 2

$\frac1{x^2}$ is not the integral of $\frac1{x^2}$.