The first experience with calculus is usually quite memorable for most of us. For some it introduces a tool of immense beauty and seemingly infinite potential to solve problems - like a super power. For others, it seems like a lifetime of suffering and madness. The reason is simple - there just too many traps and pitfalls in calculus to worry about, and even seasoned mathematicians are sometimes found scratching their heads.

In this article, we are going to see a demonstration of how things can go horribly wrong if we are not careful, while dealing with even the simplest of integrals.

# Proving that 1 is equal to 0

Consider the following simple integral -

$$\int \frac{1}{x} \cdot dx$$

To solve this, we are going to use a well known formula of integration by parts, which is given by

$$\int u~v \cdot dx = u \int v\cdot dx - \int \left(u^{\prime} \int v \cdot dx\right)dx.$$

For our case, we choose $u=1/x$ and $v=1$, so that

\begin{aligned}I =& \int \frac{1}{x} \cdot 1 \cdot dx \\ \Rightarrow I =& \frac{1}{x} \int 1\cdot dx - \int \left(-\frac{1}{x^2} \int 1 \cdot dx\right)dx \\ \Rightarrow I=& \frac{1}{x} \cdot x + \int \left(\frac{1}{x^2} \cdot x\right)dx \\ \Rightarrow I=& 1+ \int \frac{1}{x} \cdot dx \\ \Rightarrow \bcancel{I}=& 1 + \bcancel{I} \\ \Rightarrow 0=& 1 \end{aligned}

If we look at this solution casually, there seems to be nothing wrong, and yet we know it is impossible!

There is a little trap here, that we overlooked.

# Disproving that 1 is equal to 0

The problem with our solution is that we ignored one key requirement of integral calculus - We should not solve integrals without its proper limits, otherwise we might not get a sensible result.

Let us now repeat the calculations, taking some arbitrarily chosen proper limits. So, this time we have

$$I = \int_a^b \frac{1}{x} \cdot dx,$$$$so that we get,$$\begin{aligned}I &= \int_a^b \frac{1}{x} \cdot 1 \cdot dx \\ \Rightarrow I &= \left [ \frac{1}{x} \int 1\cdot dx \right]_a^b - \int_a^b \left(-\frac{1}{x^2} \int 1 \cdot dx\right)dx \\ \Rightarrow I &= \left [\frac{1}{x} \cdot x \right]_a^b + \int_a^b \left(\frac{1}{x^2} \cdot x\right)dx \\ \Rightarrow I &= _a^b+ \int_a^b \frac{1}{x} \cdot dx \\ \Rightarrow I &= [1-1] + I \\ \Rightarrow \bcancel{I} &= 0 + \bcancel{I} \\ \Rightarrow 0 &= 0 \end{aligned}

and all is well! So, the moral of the story is - Always choose some limits while doing integrals.

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