Let f(α) denote the area of the region in the first quadrant bounded by x = 0, x = 1, y² = x and y =

Q: Wait, why can we just remove the modulus signs for $|x-5|$ and $|1-x|$? Isn't that dangerous?

Great question! It is not dangerous if you respect the domain. We are integrating from $x=0$ to $x=1$. In this specific interval, $x$ is always less than $5$, so $x-5$ is always negative. Thus, $|x-5| = -(x-5) = 5-x$. Similarly, $1-x$ is always non-negative in this range. Always check your interval before opening a modulus!

Q: Do I really need to calculate two separate integrals? Is there a property of linearity I can use?

You could technically find the general expression for $f(\alpha)$ and then evaluate, but that involves piecewise integration for the general case. Since the question asks for $f(0) + f(1)$, calculating them individually is actually the most efficient and least error-prone path. Don't overcomplicate when the direct route is clear!

Q: Why is the lower boundary $\sqrt{x}$? The problem says $y^2 = x$.

Ah, the classic trap! $y^2 = x$ implies $y = \pm \sqrt{x}$. However, the problem explicitly states we are in the 'first quadrant'. In the first quadrant, $y$ must be positive, so we discard the negative root and take $y = \sqrt{x}$. Always read the quadrant constraints carefully!

Question

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^2 = x$ and $y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then $(f(0) + f(1))$ is equal to

Options:

(A) 12

(B) 14

(C) 7

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at a region trapped in the first quadrant. The boundaries are defined by the vertical lines $x=0$ and $x=1$.

The lower boundary is the curve $y^2 = x$. Since we are restricted to the first quadrant, we discard the negative root and focus on the elegant curve $y = \sqrt{x}$.

The upper boundary is defined by the function:
$$y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$$

We are tasked with finding the sum of the areas $f(0)$ and $f(1)$, corresponding to the snapshots at $\alpha = 0$ and $\alpha = 1$.

### The Calm of $\alpha = 0$

When we set $\alpha = 0$, the upper boundary function undergoes a miraculous transformation. The terms involving $\alpha$ vanish, leaving:
$$y = |0 \cdot x - 5| - |1 - 0 \cdot x| + 0 \cdot x^2 = |-5| - |1| = 5 - 1 = 4$$

The upper boundary simplifies to a flat, horizontal line at $y = 4$. The area $f(0)$ is the region between $y = 4$ and $y = \sqrt{x}$ from $x=0$ to $x=1$.

The integral is defined as:
$$f(0) = \int_{0}^{1} (4 - \sqrt{x}) \, dx$$

Evaluating this integral:
$$f(0) = \left[ 4x - \frac{2}{3}x^{3/2} \right]_{0}^{1} = 4 - \frac{2}{3} = \frac{10}{3}$$

### The Modulus Challenge at $\alpha = 1$

Now, we turn our attention to $\alpha = 1$. The upper boundary becomes $y = |x - 5| - |1 - x| + x^2$.

For the interval $x \in [0, 1]$, the expression $x - 5$ is negative, so $|x - 5| = 5 - x$. Simultaneously, $1 - x$ is non-negative, so $|1 - x| = 1 - x$.

Substituting these into the function:
$$y = (5 - x) - (1 - x) + x^2$$
$$y = 5 - x - 1 + x + x^2 = 4 + x^2$$

The upper boundary simplifies to the parabola $y = 4 + x^2$.

### The Final Synthesis

We now calculate the second integral:
$$f(1) = \int_{0}^{1} (4 + x^2 - \sqrt{x}) \, dx$$

Integrating term by term:
$$f(1) = \left[ 4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2} \right]_{0}^{1}$$
$$f(1) = 4 + \frac{1}{3} - \frac{2}{3} = 4 - \frac{1}{3} = \frac{11}{3}$$

Finally, we compute the sum $f(0) + f(1)$:
$$\frac{10}{3} + \frac{11}{3} = \frac{21}{3} = 7$$

The final result is **7**.

Animated Solution for Mathematics - Definite Integration: Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^{2} = x$ and $y = ∣ α x - 5∣ - ∣1 - α x ∣ + α x^{2}$ . Then $(f (0) + f (1))$ is equal to

Select Answer:

Visualized Solution

Understanding the Region

Defining the Area Function $f (α)$

Case 1: Setting up for $α = 0$

Simplifying $y$ for $α = 0$

Visualizing $f (0)$

Evaluating $f (0)$

Case 2: Setting up for $α = 1$

Simplifying Modulus for $x \in [0, 1]$

Final Expression for $y$ at $α = 1$

Visualizing $f (1)$

Evaluating $f (1)$

Final Calculation: $f (0) + f (1)$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Calm of $α = 0$

The Modulus Challenge at $α = 1$

The Final Synthesis

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Select Answer:

Visualized Solution

Understanding the Region

Defining the Area Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(α)

Case 1: Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=0

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=0

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

Case 2: Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=1

Simplifying Modulus for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[0,1]

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=1

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)+f(1)