For which of the following values of m, is the area of the region bounded by the curve y = x - x² an

Q: Why do we use the absolute value in the area integral?

Great question! When we set up the integral $\int_{0}^{1-m} ((1-m)x - x^2) dx$, we don't know if $1-m$ is positive or negative, or which curve is on top. The absolute value ensures that the area, which must be a positive physical quantity, remains positive regardless of the orientation of the line $y = mx$ relative to the parabola.

Q: Can we just use the shortcut formula for the area of a parabola bounded by a line?

Absolutely! The area bounded by a parabola $y = ax^2 + bx + c$ and a line $y = mx + k$ is indeed $\frac{|a|}{6}(\beta - \alpha)^3$, where $\alpha$ and $\beta$ are the roots of the intersection equation. In our case, $a = -1$, so the area is $\frac{1}{6}(\beta - \alpha)^3$. This is a powerful tool to save time in competitive exams!

Q: Why are there two values for $m$?

Because the line $y = mx$ can intersect the parabola $y = x - x^2$ at different positions depending on its slope. A positive slope $m$ might trap an area on one side of the y-axis, while a negative slope $m$ traps an area on the other side. Both configurations can yield the same area of $9/2$.

Question

For which of the following values of $m$, is the area of the region bounded by the curve $y = x - x^2$ and the line $y = mx$ equals $9/2$?

Options:

(A) -4

(B) -2

(C) 2

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The parabola is defined by the equation $y = x - x^2$. The line passing through the origin is given by $y = mx$.

We seek the values of $m$ such that the area trapped between these two curves is exactly $\frac{9}{2}$.

### Finding the Intersection: The Algebraic Foundation

To determine the boundaries of the region, we set the equations equal to each other:
$$x - x^2 = mx$$

Rearranging the terms, we obtain:
$$x^2 + (m - 1)x = 0$$

Factoring out $x$, we find:
$$x(x + m - 1) = 0$$

This yields two intersection points: $x = 0$ and $x = 1 - m$. These points serve as the limits of our integration.

### The Integral: Measuring the Void

The area $A$ is defined by the integral of the upper curve minus the lower curve. We express this as:
$$A = \left| \int_{0}^{1-m} ((1 - m)x - x^2) \, dx \right|$$

The absolute value is necessary to ensure the area remains positive, regardless of the orientation of the line relative to the parabola.

### The Calculation: Elegance in Simplification

Performing the integration term by term, we evaluate:
$$\left[ \frac{(1 - m)x^2}{2} - \frac{x^3}{3} \right]_{0}^{1-m}$$

Substituting the upper limit $x = 1 - m$ into the expression, we get:
$$\frac{(1 - m)(1 - m)^2}{2} - \frac{(1 - m)^3}{3}$$

This simplifies to:
$$\frac{(1 - m)^3}{2} - \frac{(1 - m)^3}{3} = \frac{(1 - m)^3}{6}$$

### The Final Reveal: Solving for $m$

We are given that the area is $\frac{9}{2}$. Therefore, we set up the following equation:
$$\left| \frac{(1 - m)^3}{6} \right| = \frac{9}{2}$$

Multiplying both sides by $6$, we obtain:
$$|1 - m|^3 = 27$$

Taking the cube root of both sides yields:
$$|1 - m| = 3$$

This results in two distinct cases:
1. $1 - m = 3 \Rightarrow m = -2$
2. $1 - m = -3 \Rightarrow m = 4$

The specific slopes that satisfy the condition are **$m = -2$** and **$m = 4$**.

Animated Solution for Mathematics - Definite Integration: For which of the following values of $m$ , is the area of the region bounded by the curve $y = x - x^{2}$ and the line $y = m x$ equals $9/2$ ?

Select Answer:

Visualized Solution

Visualizing the Curves

Finding Intersection Points

Solving for $x$

The Bounded Region

Setting up the Integral

Simplifying the Integrand

Performing the Integration

Applying the Limits

Simplifying the Area Expression

Equating to Given Area

Solving the Absolute Value

Finding the Values of $m$

Visualizing the Final Solutions

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Finding the Intersection

The Integral

The Calculation

The Final Reveal

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