
The correct evaluation of $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4})\lvert.\text{d}x$ is
(A) $2+\surd2$
(B) $2-\surd2$
(C) $-2+\surd2$
(D) $0$
Answer
233.1k+ views
Hint: We can determine from the given question that the function does not have any positive values across its full range. This function's value is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$. Therefore, the function is divided into two halves, one from $0$ to $\dfrac{\pi}{4}$and another from $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.
Complete step by step solution: The given integral is: I = $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4})|.\text{d}x$
Here the given function is f(x) = $\|sin~(x-\dfrac{\pi}{4})|$ and the total range over which the integral is to be done is $0$ to $\dfrac{\pi}{2}$, with lower limit $0$ and upper limit $\dfrac{\pi}{2}$. We observe that the value of this function is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.
So, the above integral can be written as I = $\int_{0}^{\dfrac{\pi}{4}}sin~(x-\dfrac{\pi}{4}).\text{d}x+\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}sin~(x-\dfrac{\pi}{4}).\text{d}x$
As integration of sin x = -cos x. So, the above integral becomes
I = $\left[-cos~(x-\dfrac{\pi}{4})\right]_{0}^{\pi/4}+\left[-cos~(x-\dfrac{\pi}{4})\right]_{\pi/4}^{\pi/2}$
We should remember that $cos 0 = 1$; cos $\dfrac{\pi}{4}$ = $\dfrac{1}{\surd2}$; cos $\dfrac{\pi}{2}$ = $0$.
Using these values to solve the question, we get
I = $(-1-(\dfrac{1}{\surd2}))+(\dfrac{1}{\surd2}-(-1))$
I = $1-\dfrac{1}{\surd2}-\dfrac{1}{\surd2}+1$
I = $2-\surd2$
Hence, the correct evaluation of $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4}).\text{d}x|$ is $2-\surd2$.
Thus, Option (B) is correct.
Note: The integration is done carefully and then the upper and lower limits are substituted carefully. Integrals with no integration limit are known as indefinite integrals. Integrals with an upper and lower limit are said to be definite integrals.
Complete step by step solution: The given integral is: I = $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4})|.\text{d}x$
Here the given function is f(x) = $\|sin~(x-\dfrac{\pi}{4})|$ and the total range over which the integral is to be done is $0$ to $\dfrac{\pi}{2}$, with lower limit $0$ and upper limit $\dfrac{\pi}{2}$. We observe that the value of this function is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.
So, the above integral can be written as I = $\int_{0}^{\dfrac{\pi}{4}}sin~(x-\dfrac{\pi}{4}).\text{d}x+\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}sin~(x-\dfrac{\pi}{4}).\text{d}x$
As integration of sin x = -cos x. So, the above integral becomes
I = $\left[-cos~(x-\dfrac{\pi}{4})\right]_{0}^{\pi/4}+\left[-cos~(x-\dfrac{\pi}{4})\right]_{\pi/4}^{\pi/2}$
We should remember that $cos 0 = 1$; cos $\dfrac{\pi}{4}$ = $\dfrac{1}{\surd2}$; cos $\dfrac{\pi}{2}$ = $0$.
Using these values to solve the question, we get
I = $(-1-(\dfrac{1}{\surd2}))+(\dfrac{1}{\surd2}-(-1))$
I = $1-\dfrac{1}{\surd2}-\dfrac{1}{\surd2}+1$
I = $2-\surd2$
Hence, the correct evaluation of $\int_{0}^{\dfrac{\pi}{2}}|sin~(x-\dfrac{\pi}{4}).\text{d}x|$ is $2-\surd2$.
Thus, Option (B) is correct.
Note: The integration is done carefully and then the upper and lower limits are substituted carefully. Integrals with no integration limit are known as indefinite integrals. Integrals with an upper and lower limit are said to be definite integrals.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Mutually Exclusive vs Independent Events: Key Differences Explained

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

Sign up for JEE Main 2026 Live Classes - Vedantu

JEE Main 2026 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

Inductive Effect and Its Role in Acidic Strength

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

