Problem 147 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\int_{1}^{4}|x-3| d x=2 A+B\), Then (A) \(A=3 / 2, B=4\) (B) \(A=1, B=1 / 2\) (C) \(A=2, B=-3 / 2\) (D) \(A=1 / 2, B=3 / 2\)

Step-by-Step Solution

Answer

Options (B) and (D) both satisfy the integral equation, but often such problems expect one unique interpretation, matching (B).

Step 1: Analyze the Absolute Value Function

The integral involves the absolute value function \(|x-3|\). This is a piecewise function: it equals \(x-3\) when \(x \geq 3\), and \(-(x-3)\) when \(x < 3\). We need to break the integral at \(x=3\), where the function changes form.

Step 2: Split the Integral into Two Intervals

Split the integral from 1 to 4 into two integrals: one from 1 to 3 and the other from 3 to 4. This gives us: \[\int_{1}^{4} |x-3| \, dx = \int_{1}^{3} (-(x-3)) \, dx + \int_{3}^{4} (x-3) \, dx.\]

Step 3: Evaluate the First Integral

Evaluate the integral \(\int_{1}^{3} (-(x-3)) \, dx\), which simplifies to \(\int_{1}^{3} (3-x) \, dx\). Calculating this gives: \[\int_{1}^{3} (3-x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{1}^{3} = (9 - 4.5) - (3 - 0.5) = 4.5 - 2.5 = 2.\]

Step 4: Evaluate the Second Integral

Evaluate the integral \(\int_{3}^{4} (x-3) \, dx\), which gives: \[\int_{3}^{4} (x-3) \, dx = \left[ \frac{x^2}{2} - 3x \right]_{3}^{4} = (8 - 12) - (4.5 - 9) = -4 - (-4.5) = 0.5.\]

Step 5: Add the Evaluated Integrals

Add the results of the two integrals from Steps 3 and 4 to find the total integral:\[\int_{1}^{4} |x-3| \, dx = 2 + 0.5 = 2.5.\]

Step 6: Compare with Given Expression

We are given that \(\int_{1}^{4} |x-3| \, dx = 2A + B\), which equals 2.5. This implies that the equation can be written as: \[2A + B = 2.5.\] From examining the options, determine which values match.

Step 7: Select the Correct Option

Check the given options:- (A) If \(A = \frac{3}{2}\) and \(B = 4\): \(2 \times \frac{3}{2} + 4 = 7\) (incorrect)- (B) If \(A = 1\) and \(B = \frac{1}{2}\): \(2 \times 1 + \frac{1}{2} = 2.5\) (correct)- (C) If \(A = 2\) and \(B = -\frac{3}{2}\): \(2 \times 2 - \frac{3}{2} = 3.5\) (incorrect)- (D) If \(A = \frac{1}{2}\) and \(B = \frac{3}{2}\): \(2 \times \frac{1}{2} + \frac{3}{2} = 2.5\) (also correct)Both (B) and (D) satisfy the equation for different interpretations of \(2A+B\).

Key Concepts

Definite Integral

In calculus, the definite integral is a fundamental concept that helps us find the area under a curve over a specific interval. The definite integral is presented in the form:

• \[ \int_{a}^{b} f(x) \, dx \]

Here, \( a \) and \( b \) are the limits between which the area under the curve \( f(x) \) is computed. In this exercise, we calculate a definite integral from 1 to 4 of the function \(|x-3|\).
To solve, the interval is split into parts where the function has different expressions, which requires understanding the nature of the function involved.

Absolute Value Function

The function absolute value, denoted by \(|x-3|\), plays a key role in our exercise. Absolute value functions display different behaviors based on the input values.
For example:

• If \( x \geq 3 \), then \(|x-3| = x-3\)
• If \( x < 3 \), then \(|x-3| = -(x-3)\)

This step is essential as it transforms a non-differentiable point into a manageable piecewise function. Understanding this function thoroughly helps split and perform integration correctly across the given interval.

Piecewise Function

A piecewise function allows us to handle situations involving functions that behave differently over different parts of their domain. In this exercise, the absolute value function \(|x-3|\) becomes:

• \(x-3\) when \(x\) is between 3 and 4
• \(3-x\) when \(x\) is between 1 and 3

By breaking the original integral into two intervals (1 to 3 and 3 to 4), we can manage each segment according to its rule system. This technique is crucial to solve the problem systematically and is a common approach in calculus to overcome variations in functions over a domain.

Integration Techniques

To find the integral of piecewise-defined functions, we employ specific integration techniques:

• Splitting the integral across relevant intervals, such as splitting \( \int_{1}^{4} |x-3| \, dx \) into \( \int_{1}^{3} (3-x) \, dx \) and \( \int_{3}^{4} (x-3) \, dx \).
• Using the fundamental theorem of calculus to evaluate each integral, followed by adding the results.

In the exercise, we separately calculate \( \int_{1}^{3} (3-x) \, dx \) to obtain 2 and \( \int_{3}^{4} (x-3) \, dx \) to obtain 0.5.
Combining these gives the total area under the curve as 2.5, showing how each integration technique brings us closer to the solution. Recognizing effective techniques ensures that even complex functions become straightforward to handle in calculus.