Let A be the area bounded by the curve y = x|x-3|, the x-axis and the ordinates x = -1 and x = 2. Th

Q: Wait, why do we split the integral at $x=0$? Can't we just integrate from $-1$ to $2$ directly?

That is a fantastic question! If you integrate directly from $-1$ to $2$, you are calculating the 'net signed area.' The part of the curve below the x-axis would cancel out the part above it. Since the question asks for the 'area bounded by the curve,' we are looking for the total geometric area, which must be positive. Splitting at the root $x=0$ allows us to treat the negative region as a positive contribution.

Q: How do I know if the function is positive or negative without graphing?

You don't need to graph it perfectly! Just pick a 'test point' in each interval. For the interval $[-1, 0]$, pick $x = -0.5$. Plugging this into $y = x(3-x)$, you get $(-0.5)(3.5) = -1.75$, which is negative. For the interval $[0, 2]$, pick $x = 1$. You get $(1)(2) = 2$, which is positive. This simple test tells you exactly when to flip the sign!

Q: I got $31/6$ for $A$, but the question asks for $12A$. Is this a trick?

It is not a trick, but it is a test of your focus! In competitive exams like JEE, the final step often requires a specific format or a multiplier. Always re-read the final line of the question before you put your pen down. You did the hard part correctly; just don't forget that final multiplication!

Question

Let $A$ be the area bounded by the curve $y = x|x-3|$, the $x$-axis and the ordinates $x = -1$ and $x = 2$. Then $12A$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The curve is defined by $y = x|x-3|$. To find the area bounded by this curve between $x = -1$ and $x = 2$, we must first resolve the absolute value.

In the interval $x \in [-1, 2]$, the value of $x$ is always less than $3$. Consequently, $x-3$ is always negative, which implies $|x-3| = -(x-3) = 3-x$.

Thus, within our interval of interest, the function simplifies to:
$$y = x(3-x) = 3x - x^2$$

### The Sign Trap: Why We Must Split

To find the total geometric area, we must account for the regions where the curve lies above or below the $x$-axis. Factoring $y = x(3-x)$, we identify roots at $x = 0$ and $x = 3$.

Within the interval $[-1, 2]$, the curve crosses the $x$-axis at $x = 0$. 
- For $x \in [-1, 0)$, the value of $y$ is negative (below the $x$-axis).
- For $x \in (0, 2]$, the value of $y$ is positive (above the $x$-axis).

The total area $A$ is given by the sum of the absolute values of the integrals:
$$A = \int_{-1}^{0} |3x - x^2| dx + \int_{0}^{2} |3x - x^2| dx$$

### The Integration Masterclass

First, we calculate $I_1 = \int_{-1}^{0} |3x - x^2| dx$. Since the function is negative here, we integrate the negated expression:
$$I_1 = \int_{-1}^{0} (x^2 - 3x) dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} ight]_{-1}^{0}$$

Evaluating at the limits:
$$I_1 = 0 - \left( \frac{(-1)^3}{3} - \frac{3(-1)^2}{2} ight) = -\left( -\frac{1}{3} - \frac{3}{2} ight) = \frac{1}{3} + \frac{3}{2} = \frac{11}{6}$$

Next, we calculate $I_2 = \int_{0}^{2} (3x - x^2) dx$. Since the function is positive here, we integrate directly:
$$I_2 = \left[ \frac{3x^2}{2} - \frac{x^3}{3} ight]_{0}^{2} = \left( \frac{3(2)^2}{2} - \frac{2^3}{3} ight) - 0 = 6 - \frac{8}{3} = \frac{10}{3}$$

Summing these components, we find the total area $A$:
$$A = \frac{11}{6} + \frac{10}{3} = \frac{11}{6} + \frac{20}{6} = \frac{31}{6}$$

### The Final Victory

The problem requires the value of $12A$. Substituting our calculated area:
$$12 	imes A = 12 	imes \frac{31}{6} = 2 	imes 31 = 62$$

Through careful analysis of the modulus and precise integration, we have arrived at the final answer: **62**.

Animated Solution for Mathematics - Definite Integration: Let $A$ be the area bounded by the curve $y = x ∣ x - 3∣$ , the $x$ -axis and the ordinates $x = - 1$ and $x = 2$ . Then $12 A$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Defining the Region

Simplifying the Modulus Function

Finding the Roots and Sign

Formulating the Area Integral

Adjusting Signs for Area

Integrating the First Part

Substituting Limits for $I_{1}$

Calculating $I_{1}$

Integrating the Second Part

Substituting Limits for $I_{2}$

Calculating $I_{2}$

Summing the Areas

Finding $A$

Calculating $12 A$

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Sign Trap

The Integration Masterclass

The Final Victory

Similar Questions

The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

The area of the region bounded by the curve $y = max {∣ x ∣, x ∣ x - 2∣}$ , then x-axis and the lines $x = - 2$ and $x = 4$ is equal to _______ .

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If the area bounded by the curve $2 y^{2} = 3 x$ , lines $x + y = 3, y = 0$ and outside the circle $(x - 3)^{2} + y^{2} = 2$ is $A$ , then $4 (π + 4 A)$ is equal to _________.

Let the area of the region ${(x, y) : ∣2 x - 1∣ \leq y \leq ∣ x^{2} - x ∣, 0 \leq x \leq 1}$ be $A$ . Then $(6 A + 11)^{2}$ is equal to ______.

The area (in sq. units) bounded by the parabola $y = x^{2} - 1$ , the tangent at the point $(2, 3)$ to it and the y-axis is :

Enter Numerical Value:

Visualized Solution

Defining the Region

Simplifying the Modulus Function

Finding the Roots and Sign

Formulating the Area Integral

Adjusting Signs for Area

Integrating the First Part

Substituting Limits for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Integrating the Second Part

Substituting Limits for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Summing the Areas

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 12A

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Sign Trap

The Integration Masterclass

The Final Victory

Similar Questions

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