If the area of the bounded region R = {(x, y) : max {0, loge x} ≤ y ≤ 2^x, 1/2 ≤ x ≤ 2} is, α(loge 2

Q: Wait, why did the integral of $\ln x$ from $\frac{1}{2}$ to $1$ disappear?

Great observation! Remember, the lower boundary is defined as $\max\{0, \ln x\}$. In the interval $[\frac{1}{2}, 1]$, the value of $\ln x$ is negative. The maximum of $0$ and a negative number is $0$. Therefore, the area under the curve in that specific interval is the area under the line $y=0$, which is zero. We only care about the region where the curve is actually positive, which starts at $x=1$!

Q: Why can't I just integrate $2^x$ as $2^x$?

That is a classic trap! Remember that the derivative of $a^x$ is $a^x \ln a$. Therefore, by the fundamental theorem of calculus, the integral must be $\frac{a^x}{\ln a}$. Always remember to divide by the natural log of the base when integrating exponential functions. It is a small detail that makes a massive difference in your final coefficient.

Q: How do I know when to stop simplifying and start comparing coefficients?

Once you have your expression in the form $A = \alpha(\ln 2)^{-1} + \beta(\ln 2) + \gamma$, you have reached the finish line. The problem gives you the structure; your job is to map your calculated constants to $\alpha$, $\beta$, and $\gamma$ by matching the terms. If your terms don't match the form, re-check your algebraic distribution and ensure you haven't missed any signs.

Question

If the area of the bounded region $R = \{(x, y) : \max \{0, \log_e x\} \leq y \leq 2^x, \frac{1}{2} \leq x \leq 2\}$ is, $\alpha(\log_e 2)^{-1} + \beta (\log_e 2) + \gamma$, then the value of $(\alpha + \beta - 2\gamma)^2$ is equal to :

Options:

(A) 8

(B) 2

(C) 4

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The region $R$ is bounded by the exponential curve $y = 2^x$ and the piecewise function $y = \max\{0, \ln x\}$.

For $x$ values in the interval $[\frac{1}{2}, 1]$, the natural logarithm $\ln x$ is negative. Consequently, the function $\max\{0, \ln x\}$ evaluates to $0$ in this range.

For $x$ values in the interval $[1, 2]$, the natural logarithm is positive, and the function follows $y = \ln x$. This transition at $x = 1$ is the critical boundary for our integration.

### The Integral Setup

The area $A$ is defined as the integral of the upper boundary minus the lower boundary from $x = \frac{1}{2}$ to $x = 2$.

Because the lower boundary is $0$ on $[\frac{1}{2}, 1]$ and $\ln x$ on $[1, 2]$, the integral simplifies to:

$$A = \int_{\frac{1}{2}}^{2} 2^x \, dx - \int_{1}^{2} \ln x \, dx$$

### The Exponential Dance

We first evaluate the integral $I_1 = \int_{\frac{1}{2}}^{2} 2^x \, dx$. Using the standard integration rule $\int a^x \, dx = \frac{a^x}{\ln a}$, we obtain:

$$I_1 = \left[ \frac{2^x}{\ln 2} \right]_{\frac{1}{2}}^{2} = \frac{2^2}{\ln 2} - \frac{2^{\frac{1}{2}}}{\ln 2}$$

This simplifies to:

$$I_1 = \frac{4 - \sqrt{2}}{\ln 2} = (4 - \sqrt{2})(\ln 2)^{-1}$$

### The Logarithmic Challenge

Next, we evaluate $I_2 = \int_{1}^{2} \ln x \, dx$. Using integration by parts, the antiderivative of $\ln x$ is $x \ln x - x$.

Evaluating this from $1$ to $2$:

$$I_2 = [x \ln x - x]_{1}^{2} = (2 \ln 2 - 2) - (1 \ln 1 - 1)$$

Since $\ln 1 = 0$, the expression simplifies to:

$$I_2 = 2 \ln 2 - 2 + 1 = 2 \ln 2 - 1$$

### The Final Convergence

We combine the results to find the total area $A$:

$$A = (4 - \sqrt{2})(\ln 2)^{-1} - (2 \ln 2 - 1)$$

Distributing the negative sign, we obtain:

$$A = (4 - \sqrt{2})(\ln 2)^{-1} - 2 \ln 2 + 1$$

Comparing this to the form $\alpha(\ln 2)^{-1} + \beta(\ln 2) + \gamma$, we identify the constants:

$$\alpha = 4 - \sqrt{2}, \quad \beta = -2, \quad \gamma = 1$$

Finally, we calculate the value of $(\alpha + \beta - 2\gamma)^2$:

$$(\alpha + \beta - 2\gamma)^2 = (4 - \sqrt{2} - 2 - 2(1))^2$$

$$(4 - \sqrt{2} - 2 - 2)^2 = (-\sqrt{2})^2 = 2$$

The final result is **2**.

Animated Solution for Mathematics - Definite Integration: If the area of the bounded region $R = {(x, y) : max {0, lo g_{e} x} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2}$ is, $α (lo g_{e} 2)^{- 1} + β (lo g_{e} 2) + γ$ , then the value of $(α + β - 2 γ)^{2}$ is equal to :

Select Answer:

Visualized Solution

Define the Bounded Region $R$

Analyze the Lower Boundary

Set up the Area Integral

Integrate the Exponential Part

Integrate the Logarithmic Part

Simplify the Logarithmic Integral

Combine to Find Total Area

Compare with Given Expression

Calculate Final Value

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Integral Setup

The Exponential Dance

The Logarithmic Challenge

The Final Convergence

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