Problem 192 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\int_{-2}^{3}\left|1-x^{2}\right| d x\) is (A) \(\frac{28}{3}\) (B) \(\frac{14}{3}\) (C) \(\frac{7}{3}\) (D) \(\frac{1}{3}\)

Step-by-Step Solution

Answer

The value is \(\frac{28}{3}\), option (A).

Step 1: Understand the Absolute Value Function

The given integral is \( \int_{-2}^{3} |1-x^2| \, dx \). We need to handle the absolute value, so we consider when \( 1 - x^2 \) is positive or negative.

Step 2: Determine Critical Points

Solve \( 1 - x^2 = 0 \) to find when the expression switches sign. This gives \( x^2 = 1 \), so \( x = 1 \) and \( x = -1 \).

Step 3: Analyze the Intervals

Examine the sign of \( 1 - x^2 \) in each interval: - \( x < -1 \): \( 1 - x^2 < 0 \) - \( -1 \leq x < 1 \): \( 1 - x^2 \geq 0 \) - \( x \geq 1 \): \( 1 - x^2 < 0 \)

Step 4: Split the Integral into Intervals

We split the integral at the critical points: \[ \int_{-2}^{3} |1 - x^2| \, dx = \int_{-2}^{-1} (-(1 - x^2)) \, dx + \int_{-1}^{1} (1 - x^2) \, dx + \int_{1}^{3} (-(1 - x^2)) \, dx \]

Step 5: Calculate Each Integral Separate

Evaluate each integral separately:- \( \int_{-2}^{-1} (-(1 - x^2)) \, dx = \int_{-2}^{-1} (x^2 - 1) \, dx \)- \( \int_{-1}^{1} (1 - x^2) \, dx \)- \( \int_{1}^{3} (-(1 - x^2)) \, dx = \int_{1}^{3} (x^2 - 1) \, dx \)

Step 6: Compute Result of Integrals

1. Compute \( \int_{-2}^{-1} (x^2 - 1) \, dx = \left[ \frac{x^3}{3} - x \right]_{-2}^{-1} = \frac{1}{3} - 1 - (\frac{-8}{3} + 2) = \frac{7}{3} \)2. Compute \( \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = 0 \)3. Compute \( \int_{1}^{3} (x^2 - 1) \, dx = \left[ \frac{x^3}{3} - x \right]_{1}^{3} = \left(\frac{27}{3} - 3\right) - \left(\frac{1}{3} - 1\right) = \frac{21}{3} + \frac{2}{3} = 7 \)

Step 7: Sum of Integral Results

Add up the results from each interval: \( \frac{7}{3} + 0 + 7 = \frac{7}{3} + \frac{21}{3} = \frac{28}{3} \)

Step 8: Choose the Correct Answer

The value of the integral \( \int_{-2}^{3} |1-x^2| \, dx \) is \( \frac{28}{3} \). Thus, the correct answer is option (A).

Key Concepts

Absolute Value Function

In calculus, the absolute value function is essential for describing the distance of numbers from zero, regardless of their sign. When working with integrals involving absolute values, like \(|1-x^2|\), we need to consider the behavior of the expression inside the absolute value. This requires analyzing when the expression, \(1 - x^2\), is positive or negative. The absolute value function affects integration because it changes the integrand based on the sign of the internal expression. Essentially:

• When the expression inside the absolute value is positive, \(|f(x)| = f(x)\).
• When it is negative, \(|f(x)| = -f(x)\).

Understanding this helps us appropriately modify intervals for integration, ensuring that we always integrate a non-negative expression.

Critical Points in Calculus

Critical points in calculus are values where the function’s behavior changes, often affecting slope and concavity. To find them for an expression like \(1 - x^2\), set it equal to zero: Find solutions by solving \( 1 - x^2 = 0 \), which gives \(x^2 = 1\), so \(x = 1\) and \(x = -1\). These values split the domain into regions where the expression \(1 - x^2\) is either positive or negative:

• For \(x < -1\), \(1 - x^2\) is negative.
• For \(-1 \leq x < 1\), \(1 - x^2\) is positive.
• For \(x \geq 1\), \(1 - x^2\) returns to being negative.

Recognizing critical points allows us to redefine the function within separate intervals, essential for evaluating integrals effectively.

Piecewise Integration

Piecewise integration is a technique used when an integrand changes behavior over different intervals due to functions like absolute values or different polynomial expressions. In our problem, since \(1 - x^2\) changes sign at critical points \(x = -1\) and \(x = 1\), we had to split the integral into three parts:\[\int_{-2}^{3} |1 - x^2| \, dx = \int_{-2}^{-1} (-(1 - x^2)) \, dx + \int_{-1}^{1} (1 - x^2) \, dx + \int_{1}^{3} (-(1 - x^2)) \, dx\]Each segment represents the expression’s adjusted behavior:

• \(x^2 - 1\) for regions where it is negative, and \(1 - x^2\) where positive.

Piecewise integration ensures we maintain the integrity of our calculations across all intervals by ensuring continuity and correctly defined functions.

Calculus Problem-Solving Steps

Approaching calculus problems systematically helps simplify complex exercises. Follow these structured steps: 1. **Understand the Problem** Start by carefully analyzing the problem. Identify what is given and what is asked. In the case of integrals, note whether absolute values are present. 2. **Identify Critical Points** Determine where the function inside the integral changes sign to find critical points, which split the integral into manageable intervals. 3. **Piecewise Integration** Apply piecewise integration by solving the integral separately over intervals formed by critical points. Treat each interval as a simpler, individual problem. 4. **Calculate and Interpret Results** Combine results from each interval. Carefully check calculations to ensure correctness and sum properly. 5. **Verify the Solution** After computing, review steps and calculations, verify with original options if given, and ensure the solution makes mathematical sense. These steps provide a solid framework for tackling calculus problems involving continuous changes in function behavior. They ensure methodical resolution and reduce errors, especially with complex integrals.