Problem 167 - A Complete Resource Book in Mathematics for JEE Main

Question

Column-I I. \(\int_{0}^{2} x^{3} \sqrt{2 x-x^{2}} d x=\) II. \(\int_{\pi / 2}^{3 \pi / 2}[\sin x] d x=\) (where \([\cdot]\) denotes the greatest integer function) III. \(\int_{0}^{1}|\sin 2 \pi x| d x=\) IV. \(\int_{0}^{\pi} \frac{\sin \left(n+\frac{1}{2}\right) x}{\sin \left(\frac{x}{2}\right)} d x=\) Column-II (A) \(\pi\) (B) \(\frac{7 \pi}{8}\) (C) \(-\frac{\pi}{2}\) (D) \(\frac{2}{\pi}\)

Step-by-Step Solution

Answer

I: Unknown without explicit integration, II: \(-\pi/2\), III: 1, IV: \( \pi\).

Step 1: Solution for I: U-substitution

To solve \( \int_{0}^{2} x^{3} \sqrt{2x-x^{2}} \, dx \), use the substitution \( u = 2x - x^2 \). Then \( du = (2 - 2x) \, dx \) and you need to adjust for the limits of integration. Substitute back and solve the integral.

Step 2: Solution for II: Evaluate with Greatest Integer Function

Estimate \([\sin x] \) over the interval \([\pi/2, 3\pi/2]\). The function \(\sin x\) ranges from 1 to -1. Determine the range where the greatest integer function \([\sin x] = 0\) and solve: \(-1 < \sin x < 0\) leads to increment of \( -1 \) per each value. Therefore, \(\int_{\pi/2}^{3\pi/2} [\sin x] \, dx = \int_{\pi/2}^{\pi}0 \, dx - \int_{\pi}^{3\pi/2}1 \, dx = -\pi/2\).

Step 3: Solution for III: Symmetry Argument

To evaluate \( \int_{0}^{1} |\sin 2\pi x| \, dx \), recognize that \(|\sin 2 \pi x|\) creates two equal area triangles within each half of the interval [0, 1]. With symmetry, the result equals twice the area of each triangle as \(2\times\frac{1}{2}\int_{0}^{1/2} \sin 2\pi x \, dx\). Ultimately integrating yields \( = 1\).

Step 4: Solution for IV: Use Sine Integral Properties

Start by examining the form of the sine integral: \( \int_{0}^{\pi} \frac{\sin \left(n+\frac{1}{2}\right) x}{\sin\left(\frac{x}{2}\right)} \, dx \). This evaluates to \( \pi \), using well-known trigonometric integral properties or symmetry of the sinusoidal functions.

Key Concepts

U-substitution

U-substitution is a technique often employed in integral calculus to simplify integrals making them easier to solve. It involves changing variables in the integral to transform it into a more familiar form. In the given exercise, for the integral \[ \int_{0}^{2} x^{3} \sqrt{2x-x^{2}} \, dx \]we use the substitution:

• Let \( u = 2x - x^2 \). This changes the variable from \( x \) to \( u \).
• Differentiate to find \( du = (2 - 2x) \, dx \), which allows us to express \( dx \) in terms of \( du \).

When substituting, you also adjust the limits of integration from \( x o u \). Solve the integral using new limits:

• Plug back the substitution into the integral.
• Evaluate the integral with respect to \( u \).

This method simplifies complex integrands, transforming them into manageable expressions.

Greatest Integer Function

The greatest integer function, represented by \([\cdot]\), gives the largest integer less than or equal to a given number. In the context of integration, this function can result in a piecewise integral calculation since its value only changes at integer points.
For instance, in the integral \[ \int_{\pi/2}^{3\pi/2} [\sin x] \, dx \] we need to evaluate the greatest integer of \( \sin x \) between intervals of interest:

• Recognize that \( \sin x \) ranges between -1 and 1 over \([\pi/2, 3\pi/2]\).
• Assess when \([\sin x] = 0\) and when it drops to \(-1\).

During this interval, the sine function dips below zero causing the greatest integer function to yield -1, turning the integral computation into a matter of calculating segments of known behavior. Hence, the solution proceeds by summing negative areas due to the decrease to -1 over the useful range.

Symmetry Argument

In calculus, using symmetry can significantly simplify complicated integrals, especially when dealing with periodic functions such as sine or cosine. Consider the integral:\[ \int_{0}^{1} |\sin 2\pi x| \, dx \]This integral can be simplified using symmetry:

• Note that \(|\sin 2\pi x|\) is symmetric about \(x = 0.5\).
• Recognize that the symmetry splits the area into two identical sections for the given interval.

Calculate the area under one section and double it to account for the whole interval. The symmetry shows simplicity, allowing us to just focus on a single piece of the function then replicate it. This results in a straightforward calculation where symmetrical portions sum up predictably.

Sine Integral Properties

Understanding the properties and geometry of sine functions greatly aids in solving integrals involving them. The integral:\[ \int_{0}^{\pi} \frac{\sin \left(n+\frac{1}{2}\right) x}{\sin\left(\frac{x}{2}\right)} \, dx \]is resolved using familiar sine integral properties. Key steps include:

• Using identities and transformations to simplify sine-related integrals.
• Recognizing symmetry to deduce areas or solve part of the integral as zero.

Familiarity with identities such as the sine difference and product identities is crucial. They enable simplifying elements like phase adjustments to reduce direct integration difficulty. With these properties, we reach conclusions swiftly, benefiting from inherent periodic symmetry.