GATE CS 2014 – Set 3 Question
Duration: 8 min
This video lesson is available to enrolled students.
AI Summary
An AI-generated summary of this video lecture.
This educational video features a calculus lecture by Yash Jain Sir, focusing on a specific definite integral problem from the GATE CS 2014 exam. The core problem is to evaluate the integral $\int_{0}^{2\pi} |x \sin x| dx$ and express the result in the form $k\pi$ to find the constant $k$. The instructor methodically breaks down the absolute value function by analyzing the sign of the sine function over the interval $[0, 2\pi]$. He splits the integral into two distinct regions where the function behaves differently, applies integration by parts to solve the resulting expressions, and carefully substitutes the limits to arrive at the final numerical answer. This problem serves as a practical application of definite integration techniques and absolute value properties.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a title card displaying "CALCULUS" amidst a background of complex mathematical formulas. The scene transitions to a whiteboard where the problem is written: "If $\int_{0}^{2\pi} |x \sin x| dx = k\pi$, then the value of $k$ is equal to ______ (GATE CS 2014 (Set 3))". The instructor begins by writing the fundamental definition of the absolute value function on the board: $|x| = egin{cases} x & x \ge 0 \ -x & x < 0 \end{cases}$. He draws a coordinate system to visualize the domain, setting the stage for analyzing the sign of the integrand.
2:00 – 5:00 02:00-05:00
The instructor analyzes the sign of $\sin x$ within the integration limits $[0, 2\pi]$. He sketches a graph of the sine wave, noting that $\sin x$ is positive in the interval $[0, \pi]$ and negative in $[\pi, 2\pi]$. Since $x$ is non-negative throughout the entire range, the absolute value $|x \sin x|$ simplifies to $x \sin x$ when $\sin x \ge 0$ and $-x \sin x$ when $\sin x < 0$. He writes this piecewise definition on the board and splits the original integral into two parts: $\int_{0}^{\pi} x \sin x dx + \int_{\pi}^{2\pi} -x \sin x dx$.
5:00 – 7:38 05:00-07:38
The instructor solves the integrals using the integration by parts formula, writing "LIATE" on the board to justify choosing $u=x$ and $dv=\sin x dx$. He derives the indefinite integral as $\int x \sin x dx = \sin x - x \cos x$. He then evaluates the definite integrals by substituting the limits. For the first part, $[\sin x - x \cos x]_0^\pi$, he substitutes $\sin \pi = 0$ and $\cos \pi = -1$ to calculate the value as $\pi$. For the second part, $[\sin x - x \cos x]_\pi^{2\pi}$, he substitutes $\sin 2\pi = 0, \cos 2\pi = 1$ and $\sin \pi = 0, \cos \pi = -1$ to calculate the value as $-3\pi$. He combines these results into the expression $\pi - (-3\pi)$, simplifying it to $4\pi$. He circles the final answer $4\pi$, indicating that $k=4$.
The lecture provides a comprehensive walkthrough of solving a definite integral with an absolute value function. The key takeaway is the technique of splitting the integration interval based on the sign changes of the function inside the absolute value. By converting the absolute value problem into a sum of standard integrals and applying integration by parts, the instructor demonstrates a clear and effective method for handling such calculus problems commonly found in engineering entrance exams.