Problem 147 - A Complete Resource Book in Mathematics for JEE Main 2017

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

Option (D) is correct: \(A=1/2, B=3/2\).

Step 1: Understand the Absolute Value

Recognize that the absolute value \(|x-3|\) changes at \(x=3\). Thus, we must consider the behavior of \(|x-3|\) in two regions: \([1,3]\) and \([3,4]\).

Step 2: Split the Integral

Since \(|x-3|\) behaves differently on each interval, we calculate \(\int_{1}^{4} |x-3| \, dx = \int_{1}^{3} (3-x) \, dx + \int_{3}^{4} (x-3) \, dx\).

Step 3: Evaluate the First Integral

Calculate \(\int_{1}^{3} (3-x) \, dx\):1. Use the antiderivative, \(3x - \frac{x^2}{2}\). 2. Evaluate from 1 to 3: \(\left[3x - \frac{x^2}{2} \right]_1^3 = (9 - \frac{9}{2}) -(3 - \frac{1}{2}) = \frac{3}{2}\).

Step 4: Evaluate the Second Integral

Calculate \(\int_{3}^{4} (x-3) \, dx\):1. Use the antiderivative, \(\frac{x^2}{2} - 3x\).2. Evaluate from 3 to 4: \(\left[ \frac{x^2}{2} - 3x \right]_3^4 = (8 - 12) - (\frac{9}{2} - 9) = \frac{1}{2}\).

Step 5: Combine Results

Add the two evaluated integrals: \(\int_{1}^{4} |x-3| \, dx = \frac{3}{2} + \frac{1}{2} = 2\).Since given \(\int_{1}^{4} |x-3| \, dx = 2A+B\), matching this to 2 gives, \(2A + B = 2\).

Step 6: Identify Correct Option

Compare with choices:(A) gives, \(2\cdot\frac{3}{2} + 4 = 2 \ 7 = 2\), no match. (B) gives, \(2\cdot1 + \frac{1}{2} = 2 \ 2.5 = 2\), no match.(C) gives, \(2\cdot2 - \frac{3}{2} = 2 \ 2.5 = 2\), no match.(D) gives, \(2\cdot\frac{1}{2} + \frac{3}{2} = 2 \ 2 = 2\), correct match.

Key Concepts

Absolute Value

The absolute value function is like a transformation that turns negative values into positive ones. For any given number or expression \(x\), the absolute value \(|x|\) refers to the distance from \(x\) to zero on the number line.

• When \(x\) is positive or zero, \(|x| = x\).
• When \(x\) is negative, \(|x| = -x\).

In our exercise, we deal with \(|x - 3|\), which means, the expression inside the absolute value, \((x - 3)\), determines its sign.
For \(x < 3\), the expression \(x - 3\) is negative, causing \(|x - 3| = -(x - 3) = 3 - x\). For \(x \geq 3\), it becomes positive, leading \(|x - 3| = x - 3\). This behavior of the absolute value divides our integral into two parts, as the function changes at \(x = 3\).

Antiderivative

Finding the antiderivative is a key step in solving integrals. An antiderivative is an operation that reverses the process of differentiation. In simple terms, it is a function whose derivative is the original function.

• The antiderivative of a constant \(a\) is \(ax + C\) where \(C\) is a constant.
• The antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), as long as \(n eq -1\).

In this problem, we split the integral into two parts: - First, for \(3 - x\), the antiderivative is \(3x - \frac{x^2}{2}\).- Second, for \(x - 3\), it becomes \(\frac{x^2}{2} - 3x\).
Both calculations are crucial as they transform our absolute value function into manageable integrals to evaluate. The antiderivative helps find the "area under the curve" analytically.

Definite Integral

A definite integral is about finding the net area between the function and the x-axis between two limits. Think of it as a way to accumulate quantities, like finding the total distance traveled over time.

• The definite integral of a function \(f(x)\) from \(a\) to \(b\) is written as \(\int_{a}^{b} f(x) \, dx\).
• It is evaluated using the antiderivative \(F(x)\) as \(F(b) - F(a)\).

In our exercise, we calculate two definite integrals due to the breaking point at \(x = 3\). For each interval, we use the corresponding antiderivative:- \(\int_{1}^{3} (3-x) \, dx\) evaluates to \(\frac{3}{2}\).- \(\int_{3}^{4} (x-3) \, dx\) evaluates to \(\frac{1}{2}\).Adding these results gives us the area (or integral) over the full range, essential for solving the problem accurately.

Calculus

Calculus is a branch of mathematics focusing on change and accumulation. It comprises two main parts: differentiation and integration. Differentiation analyzes rates of change, while integration finds accumulated quantities. In our exercise, the focus is on integration, a primary tool for calculating definite integrals to determine areas and solve complex problems.

• Differentiation helps find slopes and rates of change.
• Integration assists in finding the total impact, such as area under curves.

In solving the given problem, calculus provides the method to break down complex functions using absolute values and solve integrals by converting them into equations that are more straightforward to calculate. This simplifies understanding and working with functions across different intervals, offering precision in mathematics and solutions.