Let a be a positive real number such that ∫(0 to a) e^(x-[x]) dx = 10e - 9 where [x] is the greatest

Q: Wait, why can't we just integrate $e^{\{x\}}$ as a single function? Why do we need to split it into $n$ and $f$?

Great question! The function $\{x\}$ is discontinuous at every integer. If you try to integrate it as a single expression, you will run into trouble because the definition of $\{x\}$ changes abruptly at every integer. By splitting the integral into full periods, we handle the discontinuities systematically and turn a complex problem into a manageable sum.

Q: How can we be sure that $n$ must be 10? Is there a formal way to prove it?

Yes! Look at the equation $ne + e^f - (n + 1) = 10e - 9$. Since $0 \le f < 1$, we know $1 \le e^f < e$. This bounds the term $e^f$ between $1$ and $2.718$. By comparing the magnitude of the terms involving $e$ and the constant terms, $n=10$ is the only integer that satisfies the equation while keeping $e^f$ within its valid range.

Q: Why is the integral from $n$ to $a$ equal to the integral from $0$ to $f$?

This is the power of periodicity! Because the function $e^{\{x\}}$ repeats every 1 unit, the area under the curve between $n$ and $n+f$ is identical to the area under the curve between $0$ and $f$. It is like shifting the graph to the left by $n$ units—the shape remains unchanged.

Question

Let $a$ be a positive real number such that $\int_0^a e^{x-[x]} dx = 10e - 9$ where $[x]$ is the greatest integer less than or equal to $x$. Then $a$ is equal to :

Options:

(A) $10 - \log_e(1 + e)$

(B) $10 + \log_e 2$

(C) $10 + \log_e 3$

(D) $10 + \log_e(1 + e)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a graph of $e^{\{x\}}$. It is not a smooth, continuous curve, but rather a series of identical, rising exponential segments, each starting from $1$ and climbing toward $e$ before resetting at every integer.

This is the magic of the fractional part function, $\{x\} = x - [x]$. Because $\{x\}$ repeats its behavior every unit, the entire function $e^{\{x\}}$ inherits this periodicity with a period $T = 1$.

When we face an integral like $\int_0^a e^{\{x\}} dx$, we are essentially calculating the area under this repeating sawtooth-like wave.

### The Strategy of Decomposition

The upper limit $a$ is a positive real number, which we can elegantly decompose into its integer part $n = [a]$ and its fractional part $f = \{a\}$, where $0 \le f < 1$.

This allows us to partition our integral into two distinct regions: the area covered by $n$ full periods and the area covered by the remaining fractional piece. We write this as:

$$ \int_0^a e^{\{x\}} dx = \int_0^n e^{\{x\}} dx + \int_n^a e^{\{x\}} dx $$

The first part, $\int_0^n e^{\{x\}} dx$, is simply $n$ times the integral over a single period, $\int_0^1 e^x dx$. The second part, $\int_n^a e^{\{x\}} dx$, by the virtue of periodicity, is equivalent to $\int_0^f e^x dx$.

### The Integration

Let us perform the calculus. The integral of $e^x$ is simply $e^x$.

For the first part, we evaluate:

$$ \int_0^1 e^x dx = [e^x]_0^1 = e^1 - e^0 = e - 1 $$

Multiplying this by $n$, we get $n(e - 1)$. For the second part, we evaluate:

$$ \int_0^f e^x dx = [e^x]_0^f = e^f - e^0 = e^f - 1 $$

Combining these, our total integral becomes $n(e - 1) + (e^f - 1)$.

### The Algebraic Dance

We are given that this total area equals $10e - 9$. So, we set up our equation:

$$ n(e - 1) + (e^f - 1) = 10e - 9 $$

Expanding this, we get $ne - n + e^f - 1 = 10e - 9$, which simplifies to:

$$ ne + e^f - (n + 1) = 10e - 9 $$

Now, we compare the coefficients. If we assume $n = 10$, the equation becomes $10e + e^f - 11 = 10e - 9$. The $10e$ terms cancel out beautifully, leaving us with $e^f - 11 = -9$, or $e^f = 2$.

### The Final Reveal

We must verify our assumption. Is $e^f = 2$ consistent with our constraint $0 \le f < 1$?

Since $e^0 = 1$ and $e^1 \approx 2.718$, and $1 < 2 < 2.718$, our value $f = \ln 2$ is perfectly valid. Thus, $n = 10$ and $f = \ln 2$.

Recalling that $a = n + f$, we arrive at our final answer:

**$a = 10 + \ln 2$**

It is a beautiful result, showing how breaking a complex problem into periodic components makes the impossible, possible.

Animated Solution for Mathematics - Definite Integration: Let $a$ be a positive real number such that $\int_{0}^{a} e^{x - [x]} d x = 10 e - 9$ where $[x]$ is the greatest integer less than or equal to $x$ . Then $a$ is equal to :

Select Answer:

Visualized Solution

The Fractional Part Function ${x}$

Splitting the Upper Limit $a$

Integrating Over Full Periods

Evaluating the First Integral

Integrating the Fractional Part

Evaluating the Second Integral

Equating to the Given Value

Comparing Coefficients

Solving for $f$

Final Value of $a$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Strategy of Decomposition

The Integration

The Algebraic Dance

The Final Reveal

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Visualized Solution

The Fractional Part Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } {x}

Splitting the Upper Limit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a