Definite IntegrationeasyFree

∫ [arg|z|] dx from 0 to 100 with z ≠ 0 | JEE

JEE Maths question with a full step-by-step solution.

Question
If z≠0z\ne 0, then ∫x=0100[arg⁡∣z∣] dx\displaystyle\int_{x=0}^{100}\big[\arg|z|\big]\,dx, where [ ⋅ ][\,\cdot\,] denotes the greatest integer function, is
A00correct
B1010
C100100
Dnot defined
Solution
Step 1: Examine ∣z∣|z|. For any non-zero complex number zz, the modulus ∣z∣|z| is a positive real number, lying on the positive real axis of the complex plane. Step 2: A positive real number has argument 00, so
arg⁡∣z∣=0.\arg|z|=0.
 Step 3: Apply the greatest integer function: [arg⁡∣z∣]=[0]=0\big[\arg|z|\big]=[0]=0. This is a constant, independent of the integration variable xx. Step 4: Integrate the constant zero:
∫01000 dx=0.\int_{0}^{100}0\,dx=0.
 Correct answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Definite Integration · easy
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
Definite Integration · easy
If f(x)=∫1/xxcos⁡t2 dtf(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt for x>0x>0, then df(x)dx\dfrac{df(x)}{dx} is
Definite Integration · easy
If ∫0π/2sin⁡10x cos⁡10x dx=2−n∫0π/2(sin⁡10x+cos⁡10x) dx\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,\cos^{10}x\,dx=2^{-n}\displaystyle\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx, n∈Nn\in\mathbb{N}, then n=n=
Definite Integration · easy
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath