Problem 90 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of the integral \(\int_{0}^{41 \pi / 4}|\cos x| d x\) is (A) \(20-\frac{1}{\sqrt{2}}\) (B) \(20+\frac{1}{\sqrt{2}}\) (C) \(19+\frac{1}{\sqrt{2}}\) (D) \(19-\frac{1}{\sqrt{2}}\)

Step-by-Step Solution

Answer

The answer is (B) \(20 + \frac{1}{\sqrt{2}}\).

Step 1: Understand the Interval

The integral begins from 0 and goes to \(41\pi/4\). It's important to note that \(|\cos x|\) will have different expressions depending on whether \(\cos x\) is positive or negative in various intervals of \(x\).

Step 2: Determine Critical Points

Determine where \(\cos x = 0\) within the interval. This happens at \(x = (2n+1)\frac{\pi}{2}\). For our interval, these points are \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \cdots, 10\frac{\pi}{2}\).

Step 3: Evaluate the Function in Sub-Intervals

On the intervals \([0, \pi/2]\), \([\pi/2, 3\pi/2]\), etc., \(\cos x\) starts positive, then goes negative. Hence, use \( \cos x\) for the first half of each \(\pi\) period and \(-\cos x\) for the second half.

Step 4: Calculate the Integral for Each Sub-Interval

For each \(\pi\) interval where \(|\cos x|\) is simplified, integrate from \(0\) to \(\pi\) considering the transformation: \(\int_0^{\pi} |\cos x| \, dx = 2\left[1\right] = 2\).

Step 5: Calculate Total Number of Cycles

From \(0\) to \(41\pi/4\), count the complete cycles of length \(\pi\). This is \(10\) complete cycles, contributing \(20\) to the integral value.

Step 6: Add Partial Cycle Contribution

After \(10\cdot\pi = 40\cdot\pi/4\), there remains a segment from \(40\pi/4\) to \(41\pi/4\), within which \(\cos\) is positive, contributing \(\sin(x)\) evaluated from \(0\) to \(\pi/4\). Compute \(\sin(\pi/4) = \frac{1}{\sqrt{2}}\).

Step 7: Final Step: Sum All Parts to Get the Answer

The total integral becomes \(20 + \frac{1}{\sqrt{2}}\), which corresponds to option (B).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in calculus, especially when dealing with integrals involving \(\sin x\) and \(\cos x\). These functions relate the angles of a triangle to the ratios of its sides. In the context of this exercise, \(\cos x\) varies between -1 and 1 as \(x\) cycles through its periods. Understanding how it behaves over each interval (typically \([0, \pi]\)) is crucial for evaluating integrals where absolute values are involved.

• The \(\cos x\) function identifies the horizontal aspect of the angle's projection in a unit circle.
• It has periodic zeros at \(x = (2n+1)\frac{\pi}{2}\), where it changes from positive to negative values or vice versa.
• This behavior creates a "wave" that repeats every \(2\pi\) radians.

Understanding this periodic nature aids in breaking down complex integrals into manageable pieces.

Absolute Value

When a function like \(\cos x\) takes both positive and negative values, absolute values come into play to keep the output non-negative. The absolute value function, denoted as \(|f(x)|\), essentially "flips" the negative portions of the function above the x-axis. In our exercise, this concept helps when separating the intervals based on the positivity of \(\cos x\).

• For \( ext{positive intervals: } |\cos x| = \cos x\).
• For \( ext{negative intervals: } |\cos x| = -\cos x\).

Understanding how to handle absolute values is crucial for ensuring that all results remain positive, as required in calculations involving magnitudes or real-world problems where negative values don't make sense. Appreciating the split into positive and negative sub-intervals ensures the integration respects the function's oscillatory nature.

Integration by Parts

Although not directly used in this exercise, integration by parts is a technique that can often help when dealing with mixed expressions. Derived from the product rule for differentiation, it is a method to transform the integral of a product of functions into potentially simpler terms. The formula used is: \[ \int u \, dv = uv - \int v \, du \]Where you pick parts of the integrand as \(u\) and \(dv\), differentiating and integrating these respectively. This concept is essential in calculus as it allows for the simplifying of many complex integrals, which could otherwise be quite involved.

Critical Points in Calculus

Critical points in calculus refer to places where a function's derivative is either zero or undefined, indicating potential maxima, minima, or inflection points. For \(\cos x\), critical points are critical for determining where the function changes sign. They indicate the points at which bending or significant behavior change occurs. In the context of this exercise:

• Critical points are each \(\frac{\pi}{2}\) plus multiples thereof, where \(\cos x = 0\).
• They split the \(x\) axis into sections where the sign of the \(\cos x\) function switches.

Recognizing these points helps in integrating trigonometric functions, as it informs the sections where considerations like absolute values must alter the integrand.