Animated Solution for Mathematics - Definite Integration: Consider the integral $I={\int }_0^{10}\frac{[x]e^{[x]}}{e^{x-1}}dx$ where $[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to:

Unlocking the Greatest Integer Function

Welcome, future engineer. Today, we are going to demystify a problem that often intimidates students: an integral involving the Greatest Integer Function, $[x]$.

When you see those square brackets, do not panic. Instead, see them as a signal to pause and visualize. The function $[x]$ is not a smooth, flowing curve; it is a staircase.

It stays flat at an integer value $n$ for the entire interval $[n,n+1)$, and then it jumps abruptly to $n+1$. This discontinuity is exactly why we cannot use standard integration techniques directly. We must break the problem down into the 'steps' of the staircase.

Taming the Exponential Beast

Before we address the staircase, let us clean up the algebra. We are given the integral:

That denominator looks messy, but remember your exponent rules. We can rewrite $\frac{1}{e^{x-1}}$ as $e^{-(x-1)}$, which is $e^{1-x}$.

Now, our integrand becomes $[x]e^{[x]}{e}^{1-x}$. By combining the exponential terms, we get $[x]e^{[x]-x+1}$.

Suddenly, the expression is much more approachable. We have transformed a complex fraction into a product of a piecewise constant function and a smooth exponential function.

The Power of Decomposition

Since $[x]$ changes its value at every integer, we must split our integral from $0$ to $10$ into unit intervals: $[0,1),[1,2),\dots ,[9,10)$. In any interval $[n,n+1)$, the value of $[x]$ is simply the constant $n$.

This allows us to rewrite our integral as a summation:

This is the heart of the solution. We have turned one intimidating integral into a sum of ten smaller, simpler ones.

And here is a delightful surprise: when $n=0$, the entire term becomes $0$. We can ignore the first interval entirely and start our sum from $n=1$.

Since $n$ is constant with respect to $x$, we pull it out of the integral:

The Final Integration

Now, we evaluate the integral ${\int }_n^{n+1}{e}^{n+1-x}dx$. The antiderivative of $e^{a-x}$ is $-e^{a-x}$.

Applying this, we get:

Evaluating at the upper limit $n+1$, we get $-e^{n+1-(n+1)}=-e^0=-1$. At the lower limit $n$, we get $-e^{n+1-n}=-e^1=-e$.

Subtracting the lower limit from the upper limit gives $(-1)-(-e)=e-1$. Notice the elegance here: the result of the integral is $e-1$, which is completely independent of $n$.

The Grand Summation

We are left with $I=(e-1){\sum }_{n=1}^{9}n$. The sum of the first nine natural numbers is a classic result:

Thus, our final answer is $45(e-1)$.

You see? By breaking the problem into logical steps, we turned a terrifying integral into a simple summation. Keep this technique in your toolkit—whenever you see $[x]$, think of the staircase, split the interval, and conquer.