If the value of the integral ∫₀⁵ (x+[x])/e^(x-[x]) dx = αe⁻¹ + β, where α, β ∈ R, 5α + 6β = 0, and [

Q: Why can't I just integrate this directly? It looks like a standard function.

Ah, the classic trap! The greatest integer function $[x]$ is not differentiable at integers and is piecewise constant. If you try to integrate it as a single expression, you are ignoring the 'jumps' in the function. You must treat it as a piecewise function, integrating over each interval where $[x]$ is constant.

Q: Is there a shortcut for the summation part?

While there isn't a single 'magic' formula, the linearity of the integral is your best friend. By swapping the summation and the integral, you reduce the complexity from five separate integrations to one single integration of a sum. Always look for ways to simplify the expression *before* you start the heavy calculus.

Q: Why do we use integration by parts here?

When you see a product of an algebraic function (like $t+4$) and an exponential function (like $e^{-t}$), integration by parts is your most powerful tool. It allows you to reduce the degree of the algebraic term, making the integral solvable. Remember the LIATE rule—algebraic functions come before exponential ones, so let $u = t+4$.

Question

If the value of the integral $\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} dx = \alpha e^{-1} + \beta$, where $\alpha, \beta \in \mathbf{R}$, $5\alpha + 6\beta = 0$, and $[x]$ denotes the greatest integer less than or equal to $x$; then the value of $(\alpha + \beta)^2$ is equal to :

Options:

(A) 100

(B) 25

(C) 16

(D) 36

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The greatest integer function, denoted as $[x]$, behaves like a staircase. It remains constant between integers and leaps at every integer value.

When evaluating the integral

$$I = \int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} dx$$

we must recognize that the function is piecewise constant. We cannot treat it as a single monolithic entity; instead, we must partition the interval $[0, 5]$ into unit steps: $[0, 1), [1, 2), [2, 3), [3, 4),$ and $[4, 5)$.

### The Strategy of Partitioning

In each interval $[k, k+1)$, the value of $[x]$ is exactly $k$. This allows us to decompose the integral into a summation of five manageable parts:

$$I = \sum_{k=0}^{4} \int_{k}^{k+1} \frac{x+k}{e^{x-k}} dx$$

This transformation turns an intimidating expression into a structured sum of simpler integrals.

### The Beauty of Substitution

To simplify the integrand, we perform the substitution $t = x - k$, which implies $dx = dt$. The limits of integration transform from $[k, k+1]$ to $[0, 1]$.

The numerator $x+k$ becomes $(t+k)+k = t+2k$, and the denominator becomes $e^t$. The integral expression now becomes:

$$I = \sum_{k=0}^{4} \int_{0}^{1} (t+2k) e^{-t} dt$$

By choosing this substitution, we have successfully aligned all five intervals to the same range $[0, 1]$, making the variable $t$ independent of the index $k$.

### The Art of Summation

Because the integral and the finite sum are linear operators, we can swap their order. We pull the summation inside the integral:

$$I = \int_{0}^{1} e^{-t} \left( \sum_{k=0}^{4} (t+2k) \right) dt$$

Calculating the inner sum, we find $\sum_{k=0}^{4} t = 5t$ and $\sum_{k=0}^{4} 2k = 2(0+1+2+3+4) = 20$. The integral simplifies to:

$$I = \int_{0}^{1} (5t+20) e^{-t} dt = 5 \int_{0}^{1} (t+4) e^{-t} dt$$

### The Final Integration

We solve the remaining integral using integration by parts, where $\int u dv = uv - \int v du$. Let $u = t+4$ and $dv = e^{-t} dt$, which gives $du = dt$ and $v = -e^{-t}$.

The antiderivative is:

$$-(t+4)e^{-t} - \int -e^{-t} dt = -(t+4)e^{-t} - e^{-t} = -e^{-t}(t+5)$$

Evaluating this from $0$ to $1$:

$$I = 5 \left[ -e^{-t}(t+5) \right]_{0}^{1} = 5 \left( (-6e^{-1}) - (-5) \right) = 25 - 30e^{-1}$$

Comparing this to the form $\alpha e^{-1} + \beta$, we identify $\alpha = -30$ and $\beta = 25$. The final result is:

$$(\alpha + \beta)^2 = (-30 + 25)^2 = (-5)^2 = **25**$$

Select Answer:

Visualized Solution

Analyzing the Integrand $I$

Breaking the Interval $[0, 5]$

Substitution $x - k = t$

Transforming the Integral

Summing the Integrands

Simplified Integral Form

Integration by Parts

Applying the Formula

Evaluating Definite Integral

Comparing Coefficients $α$ and $β$

Verifying $5 α + 6 β = 0$

Final Calculation of $(α + β)^{2}$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Strategy of Partitioning

The Beauty of Substitution

The Art of Summation

The Final Integration

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Select Answer:

Visualized Solution

Analyzing the Integrand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Breaking the Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,5]

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−k=t