Additivity over Intervals

To solve this, we must determine the value of $$f(x) = [\sin x]$$ by looking at the range of $$\sin x$$ within the four quadrants.

1. Range Analysis of $$[\sin x]$$

The Greatest Integer Function $$[y]$$ "rounds down" to the nearest integer.

• In the interval $$[0, \pi]$$ (Quadrants I and II):
  • For most values, $$0 \le \sin x < 1$$, which means $$[\sin x] = 0$$.
  • At exactly $$x = \frac{\pi}{2}$$, $$\sin x = 1$$, so $$[1] = 1$$.
  • Effect on Integral: Since a single point ($$x = \frac{\pi}{2}$$) has no "width," it does not change the area under the curve. Thus, the integral from $$0$$ to $$\pi$$ is $$0$$.
• In the interval $$(\pi, 2\pi)$$ (Quadrants III and IV):
  • Here, $$\sin x$$ is negative. Specifically, $$-1 \le \sin x < 0$$.
  • For all these values, the "floor" or greatest integer is $$-1$$.
  • Even at $$x = \frac{3\pi}{2}$$ where $$\sin x = -1$$, the value of $$[-1]$$ remains $$-1$$.
  • Effect on Integral: This creates a rectangle with a height of $$-1$$ and a width of $$\pi$$.

2. Computing the Area (Integral)

We can now split the integral at the point where the behavior changes:

$$I = \int_{0}^{\pi} [\sin x] \, dx + \int_{\pi}^{2\pi} [\sin x] \, dx$$

Substituting the values we found:

$$I = \int_{0}^{\pi} (0) \, dx + \int_{\pi}^{2\pi} (-1) \, dx$$

Evaluating the second part:

$$I = 0 + \left[ -x \right]_{\pi}^{2\pi}$$ $$I = -(2\pi) - (-\pi)$$  $$I = -2\pi + \pi = -\pi$$

1. Analyze the Integrand: Let $$f(x) = \frac{1}{e^{x-1}} = e^{1-x}$$. The Greatest Integer Function $$[e^{1-x}]$$ is non-zero only when $$e^{1-x} \ge 1$$.
2. Determine the Relevant Interval:  $$e^{1-x} \ge 1$$ implies $$1-x \ge 0$$, or $$x \le 1$$.
   • For $$x > 1$$, the integrand becomes $$0$$. Thus, the effective integration limits are $$0$$ to $$1$$.
3. Split the Integral at Integer Boundaries:
   • In the interval $$x \in [0, 1]$$, the function $$e^{1-x}$$ ranges from $$e^1$$ ($$\approx 2.718$$) down to $$e^0 = 1$$.
   • The function hits the integer value 2 when $$e^{1-x} = 2 \implies 1-x = \ln 2 \implies x = 1 - \ln 2$$.
   • From $$x = 0$$ to $$1 - \ln 2$$: $$2 \le e^{1-x} < e$$, so $$[e^{1-x}] = 2$$.
   • From $$x = 1 - \ln 2$$ to $$1$$: $$1 \le e^{1-x} < 2$$, so $$[e^{1-x}] = 1$$.
4. Evaluate the Integral:$$\int_{0}^{1-\ln 2} 2 \, dx + \int_{1-\ln 2}^{1} 1 \, dx$$ $$= 2(1 - \ln 2 - 0) + 1(1 - (1 - \ln 2))$$ $$= 2 - 2\ln 2 + \ln 2 = 2 - \ln 2$$
5. Find $$\alpha$$ and $$\alpha^3$$:
   • Comparing $$2 - \ln 2$$ with $$\alpha - \log_e 2$$, we find $$\alpha = 2$$.
   • $$\alpha^3 = 2^3 = \mathbf{8}$$.

Correct Option: B (8)

Step 1: Determine the explicit form of f(x)

We are given the functional equation:

$$f(x + 1) = e \cdot f(x)$$

Since f(0) = 1, we can observe the progression for integer values:

$$f(1) = e \cdot f(0) = e^1$$

$$f(2) = e \cdot f(1) = e^2$$

For any x belonging to R, the continuous differentiable function that satisfies this scaling relation along with the initial condition is:

$$f(x) = e^x$$

Substituting this into our target integral gives:

$$I = \int_{0}^{2} [x] \cdot x \cdot e^x \, dx$$

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Step 2: Interval Splitting for the Piecewise Component [x]

The greatest integer function [x] changes value at integer steps. We split the integration limits from 0 to 2 into two sub-intervals: [0, 1) and [1, 2].

$$I = \int_{0}^{1} [x] \cdot x \cdot e^x \, dx + \int_{1}^{2} [x] \cdot x \cdot e^x \, dx$$

Over these intervals, the value of [x] is:

For x in [0, 1), [x] = 0

For x in [1, 2), [x] = 1

Substituting these values into the integrals:

$$I = \int_{0}^{1} (0) \cdot x \cdot e^x \, dx + \int_{1}^{2} (1) \cdot x \cdot e^x \, dx$$

$$I = 0 + \int_{1}^{2} x e^x \, dx$$

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Step 3: Apply Integration by Parts

To evaluate the remaining definite integral, we use integration by parts:

$$\int u \, dv = uv - \int v \, du$$

Let:

$$u = x \implies du = dx$$

$$dv = e^x \, dx \implies v = e^x$$

Applying the boundaries:

$$\int_{1}^{2} x e^x \, dx = \left[ x e^x \right]_{1}^{2} - \int_{1}^{2} e^x \, dx$$

Evaluating the first part at the upper and lower limits:

$$\left[ x e^x \right]_{1}^{2} = (2 \cdot e^2) - (1 \cdot e^1) = 2e^2 - e$$

Evaluating the second part:

$$\int_{1}^{2} e^x \, dx = \left[ e^x \right]_{1}^{2} = e^2 - e$$

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Step 4: Combine the Evaluated Terms

Subtracting the second term from the first term gives:

$$I = (2e^2 - e) - (e^2 - e)$$

$$I = 2e^2 - e - e^2 + e$$

$$I = e^2$$