Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

The area of the region bounded by the curve y = maxleft\|x|, x|x - 2|right\ [cite: 699], the x-axis and the lines x = -2 and x = 4 is equal to[cite: 699, 702]:

Numerical Answer Type:
Enter a numerical value Answer: 12 to 12 +4 marks

Solution & Explanation

### Related Formula Definite integration geometry area: Split boundary zones around intersections where functional dominant switches occur.
Area Under Bounded Curves diagram for Q75 - JEE Main 2025 Morning
Area Under Bounded Curves diagram for Q75 - JEE Main 2025 Morning
### Core Logic Analyze intersection points between y_1 = |x| and y_2 = x|x-2| across required integration span regions: - For x in [-2, 0]: |x| = -x and x|x-2| = -x(2-x) = x^2 - 2x. Max curve tracks through distinct segments. - For positive sectors, compute intersections: x = x(2-x) implies x = 1 or x=0. Also check switch locations where graphs swap dominance. ### Step 1: Setting up separate area integral blocks Using geometric area partitions calculated across continuous regions [cite: 1495]: textArea = frac12 times 2 times 2 + frac12 times 3 times 3 + frac12 times 1 times 11 = 12 [cite: 1495] Alternatively, splitting the boundary metrics via continuous definite limits yields identical whole tracking blocks matching exactly to 12 total units[cite: 1495]. ### Pattern Recognition Plotting multiple curves dynamically highlights dominance shifts quickly. Computing distinct straight triangular chunks saves valuable integration time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals (Application of Integrals)

Reference Study Guides

Integrals

More Integrals Previous-Year Questions

Q71 2025 Definite Integration with GIF
Let [cdot] denote the greatest integer function. If int_0^e^3left[frac1e^x - 1right]mathrmdx = alpha -log_e2, then alpha^3 is equal to ________.
Numerical Answer. Answer: 8 to 8

Solution

### Related Formula Greatest Integer Function boundaries: [f(x)] = k quad textfor quad k le f(x) < k+1, \ k in mathbbZ ### Core Logic Analyze the value variations of f(x) = e^1-x across the integration limits [0, e^3] to break down the integral into distinct piecewise continuous intervals. ### Step 1: Determine Step Function Transition Points Let y = e^1-x. * At x = 0 implies y = e^1 approx 2.718 * As x increases, e^1-x decreases monotonically. * Find x where y = 2 implies e^1-x = 2 implies 1-x = ln 2 implies x = 1 - ln 2. * Find x where y = 1 implies e^1-x = 1 implies 1-x = 0 implies x = 1. * At the final boundary x = e^3 implies y = e^1-e^3, which is a very small positive decimal strictly inside (0,1). ### Step 2: Split the Definite Integral Rewrite the integral based on the isolated interval blocks: I = int_0^1-ln 2 2 \, mathrmdx + int_1-ln 2^1 1 \, mathrmdx + int_1^e^3 0 \, mathrmdx ### Step 3: Perform Integrations I = 2[x]_0^1-ln 2 + 1[x]_1-ln 2^1 + 0 I = 2(1 - ln 2 - 0) + 1(1 - (1 - ln 2)) = 2 - 2ln 2 + ln 2 = 2 - ln 2 ### Step 4: Solve for Alpha Cubed Compare the integrated value to alpha - ln 2: alpha - ln 2 = 2 - ln 2 implies alpha = 2 alpha^3 = 2^3 = 8 ### Pattern Recognition Always map the function values at the extreme boundary points first. Tracking the downward path from 2.71 rightarrow 2 rightarrow 1 rightarrow 0 reveals exactly where the integer thresholds are crossed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals
Q59 2025 Definite Integrals
The integral int_0^pi frac8x \, dx4cos^2 x + sin^2 x is equal to
  • A. 2pi^2
  • B. 4pi^2
  • C. pi^2
  • D. frac3pi^22

Solution

### Related Formula Using the properties of definite integrals: int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx Also, if f(2a-x) = f(x), then: int_0^2a f(x) \, dx = 2 int_0^a f(x) \, dx ### Core Logic Let: I = int_0^pi frac8x \, dx4cos^2 x + sin^2 x quad text--- (1) Applying x to pi-x: I = int_0^pi frac8(pi - x) \, dx4cos^2 x + sin^2 x quad text--- (2) ### Step 1: Eliminating the x term in numerator Adding (1) and (2): 2I = 8pi int_0^pi fracdx4cos^2 x + sin^2 x I = 4pi int_0^pi fracdx4cos^2 x + sin^2 x Since the integrand is symmetric about x = pi/2: I = 8pi int_0^pi/2 fracdx4cos^2 x + sin^2 x ### Step 2: Integration using sec^2 x substitution Divide numerator and denominator by cos^2 x: I = 8pi int_0^pi/2 fracsec^2 x \, dx4 + tan^2 x Let t = tan x implies dt = sec^2 x \, dx At x = 0, t = 0; at x = pi/2, t to infty. I = 8pi int_0^infty fracdtt^2 + 2^2 I = 8pi left[ frac12 tan^-1left(fract2right) right]_0^infty = 4pi left( fracpi2 - 0 right) = 2pi^2 ### Pattern Recognition The presence of a linear x factor in the numerator of a definite integral with symmetric trigonometric bounds is almost always eliminated using the a+b-x property. This reduces the integral to a standard substitution problem. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals
Q65 2025 Methods of Integration by Substitution
Let f(x) = int x^3sqrt3 - x^2 \, mathrmdx[cite: 635, 637]. If 5f(sqrt2) = -4 [cite: 638], then f(1) is equal to[cite: 642]:
  • A. -frac2sqrt25
  • B. -frac8sqrt25
  • C. -frac4sqrt25
  • D. -frac6sqrt25

Solution

### Related Formula Method of algebraic parameter substitution: Set 3-x^2 = t^2 implies -2xmathrmdx = 2tmathrmdt implies xmathrmdx = -tmathrmdt ### Core Logic Perform the specified variable parameter replacement steps [cite: 1361, 1362]: 3 - x^2 = t^2 implies x \, mathrmdx = -t \, mathrmdt [cite: 1361, 1362] Rewrite the internal integral block components [cite: 1363]: f(x) = int x^2 cdot sqrt3-x^2 cdot (x \, mathrmdx) = int (3-t^2) cdot t cdot (-t \, mathrmdt) [cite: 1363] = int (t^4 - 3t^2) \, mathrmdt = fract^55 - t^3 + C [cite: 1363, 1366] Return to original reference variable x [cite: 1366]: f(x) = frac(3-x^2)^5/25 - (3-x^2)^3/2 + C [cite: 1366] ### Step 1: Constant integration resolving Evaluate function boundary conditions at x = sqrt2 [cite: 1366]: f(sqrt2) = frac(3-2)^5/25 - (3-2)^3/2 + C = frac15 - 1 + C = -frac45 + C [cite: 1366] Given 5f(sqrt2) = -4 implies f(sqrt2) = -frac45 [cite: 638, 1366]. -frac45 + C = -frac45 implies C = 0 [cite: 1366] ### Step 2: Numeric tracking value Evaluate final targeted definition state value at x=1 [cite: 1367]: f(1) = frac(3-1)^5/25 - (3-1)^3/2 = frac2^5/25 - 2^3/2 [cite: 1367] = 2^3/2left(frac25 - 1right) = 2sqrt2left(-frac35right) = -frac6sqrt25 [cite: 1367, 1368] ### Pattern Recognition Splitting powers of x to create a direct match with internal derivative differential flags speeds up the integration transformation sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals
Q67 2025 Definite Integral of Greatest Integer Function
Let the domain of the function f(x) = log_2log_4log_6(3 + 4x - x^2) be (a, b)[cite: 663]. If int_0^b-a[x^2]dx = p - sqrtq - sqrtr [cite: 664], where p, q, r in mathbbN [cite: 664] and gcd(p, q, r) = 1 [cite: 668], and [cdot] represents the greatest integer function [cite: 668], then p + q + r is equal to[cite: 668]:
  • A. 10
  • B. 8
  • C. 11
  • D. 9

Solution

### Related Formula Domain of log chain iterations: For log_2log_4(M) > 0, we require log_4(M) > 1 implies M > 4. ### Core Logic Trace internal arguments outward sequentially [cite: 1393, 1394]: log_4log_6(3 + 4x - x^2) > 0 implies log_6(3 + 4x - x^2) > 1 [cite: 1393, 1394] 3 + 4x - x^2 > 6^1 implies x^2 - 4x + 3 < 0 [cite: 1395, 1396] (x-1)(x-3) < 0 implies x in (1, 3) [cite: 1397, 1398] Thus, determine limits [cite: 1399]: a = 1, quad b = 3 implies b - a = 2 [cite: 1399] ### Step 1: Setting up the greatest integer function integration We need to evaluate int_0^2 [x^2] \, mathrmdx[cite: 1400]. Identify step boundary switch locations inside range [0, 2] [cite: 1400]: - For x in [0, 1): [x^2] = 0 - For x in [1, sqrt2): [x^2] = 1 - For x in [sqrt2, sqrt3): [x^2] = 2 - For x in [sqrt3, 2): [x^2] = 3 Set up separate boundary component integrations [cite: 1400]: int_0^2 [x^2] \, mathrmdx = int_0^1 0 \, mathrmdx + int_1^sqrt2 1 \, mathrmdx + int_sqrt2^sqrt3 2 \, mathrmdx + int_sqrt3^2 3 \, mathrmdx [cite: 1400] = 0 + (sqrt2 - 1) + 2(sqrt3 - sqrt2) + 3(2 - sqrt3) [cite: 1400] = sqrt2 - 1 + 2sqrt3 - 2sqrt2 + 6 - 3sqrt3 = 5 - sqrt2 - sqrt3 [cite: 1400] ### Step 2: Matching coefficients Compare values with requested answer template shape [cite: 1400]: 5 - sqrt2 - sqrt3 = p - sqrtq - sqrtr [cite: 1400] p = 5, quad q = 2, quad r = 3 [cite: 1400] textFinal Sum = p + q + r = 5 + 2 + 3 = 10 [cite: 1400] ### Pattern Recognition Integrals over greatest integer configurations change value exactly where the inner expression tracks through integer milestones. Mapping boundaries accurately resolves calculations smoothly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals

More Integrals Questions — jee_main_2025_03_april_morning

Practice all Integrals previous-year questions →

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics
Latest from the Blog
View all →

Loading articles...