6int_0^pileft|left(sin3x+sin2x+sin xright)right|dx is equal to.....

Numerical Answer Type:
Enter a numerical value Answer: 17 to 17 +4 marks

Solution & Explanation

### Related Formula sin A + sin B = 2 sinleft(fracA+B2right) cosleft(fracA-B2right) sin 2x = 2 sin x cos x cos 2x = 2 cos^2 x - 1 ### Core Logic Let I = 6int_0^pi|sin 3x + sin x + sin 2x| dx. Apply sum-to-product on sin 3x + sin x: sin 3x + sin x = 2 sin(2x) cos(x) So the expression becomes: |2 sin(2x) cos x + sin 2x| = |sin 2x (2 cos x + 1)| = |2 sin x cos x (2 cos x + 1)| Since x in [0, pi], sin x geq 0. We can pull it out of the modulus. I = 12 int_0^pi sin x |2 cos^2 x + cos x| dx ### Step 1: Coordinate Substitution Substitute t = cos x, then dt = -sin x dx. Limits: when x = 0, t = 1. When x = pi, t = -1. I = 12 int_-1^1 |2t^2 + t| dt ### Step 2: Resolve Modulus Intervals The roots of 2t^2 + t = 0 are t = 0 and t = -1/2. The quadratic 2t^2 + t is negative in the interval (-1/2, 0) and positive elsewhere. Split the integral: I = 12 left[ int_-1^-1/2 (2t^2 + t) dt - int_-1/2^0 (2t^2 + t) dt + int_0^1 (2t^2 + t) dt right]
Integration of modulus polynomial function diagram for Q25 - JEE Main 2026 Morning
Integration of modulus polynomial function diagram for Q25 - JEE Main 2026 Morning
### Step 3: Evaluate Integrals Anti-derivative: F(t) = frac2t^33 + fract^22. F(1) = 2/3 + 1/2 = 7/6 F(0) = 0 F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24 F(-1) = -2/3 + 1/2 = -1/6 Evaluate each segment: 1) int_-1^-1/2 = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24 2) -int_-1/2^0 = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24 3) int_0^1 = F(1) - F(0) = 7/6 - 0 = 28/24 Sum of parts inside bracket: frac524 + frac124 + frac2824 = frac3424 = frac1712 ### Step 4: Final Output I = 12 times frac1712 = 17 ### Pattern Recognition Whenever an integral features a cascading sum of sine frequencies like sin(kx), pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under t = cos x. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

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Definite Integration

More Definite Integration Previous-Year Questions — Page 2

Q71 jee_main_2025_29_jan_evening Integration of Modulus and Greatest Integer Functions
If 24int_0^fracpi4left(sin left|4x - fracpi12right| + [2sin x]right)mathrmdx = 2pi +alpha, where [cdot ] denotes the greatest integer function, then alpha is equal to
Numerical Answer. Answer: 12 to 12

Solution

### Related Formula Additivity property of definite integrals over split interval boundary points: int_a^c g(x) \, dx = int_a^b g(x) \, dx + int_b^c g(x) \, dx ### Core Logic Separate the integration tracking flow into two component operations: I_1 = int_0^fracpi4 sinleft|4x - fracpi12 ight| \, dx I_2 = int_0^fracpi4 [2sin x] \, dx ### Step 1: Solve Modulus Integral Term The argument changes sign inside absolute value bounds when 4x - fracpi12 = 0 implies x = fracpi48: I_1 = int_0^fracpi48 -sinleft(4x - fracpi12 ight) \, dx + int_fracpi48^fracpi4 sinleft(4x - fracpi12 ight) \, dx = frac14 left[ cosleft(4x - fracpi12 ight) right]_0^fracpi48 - frac14 left[ cosleft(4x - fracpi12 ight) right]_fracpi48^fracpi4 Evaluating across numerical boundaries gives: 24 cdot I_1 = 24left(frac12right) = 12 ### Step 2: Solve Greatest Integer Component and Combine Analyze step limits inside greatest integer block [2sin x]: For x in left[0, fracpi6right), 0 le 2sin x < 1 implies [2sin x] = 0. For x in left[fracpi6, fracpi4right], 1 le 2sin x < sqrt2 implies [2sin x] = 1. I_2 = int_0^fracpi6 0 \, dx + int_fracpi6^fracpi4 1 \, dx = fracpi4 - fracpi6 = fracpi12 Combine both sections aggregated by multiplier 24: textTotal = 12 + 24left(fracpi12 ight) = 2pi + 12 Comparing directly with expression statement parameters 2pi + alpha isolates response value: alpha = 12 ### Pattern Recognition Modulus arguments and step functions require isolating inflection transition numbers directly to break integrations up neatly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration
Q63 jee_main_2025_28_jan_morning Properties of Definite Integrals (King's Property)
If int_-fracpi2^fracpi2frac96x^2cos^2x(1 + e^x) dx = pi (alpha pi^2 +beta),alpha ,beta in mathbbZ, then (alpha + beta)^2 equals: (1) 144 (2) 196 (3) 100 (4) 64
  • A. 144
  • B. 196
  • C. 100
  • D. 64

Solution

### Related Formula King's property for definite integration: int_a^b f(x) dx = int_a^b f(a+b-x) dx ### Core Logic Apply the identity x to -x to the integral: I = int_-fracpi2^fracpi2 frac96x^2 cos^2 x1 + e^x dx = int_-fracpi2^fracpi2 frac96x^2 cos^2 x1 + e^-x dx ### Step 1: Adding both integral variations Adding the equations eliminates the exponential denominator term (1+e^x): 2I = int_-fracpi2^fracpi2 96x^2 cos^2 x cdot left[frac11+e^x + frace^x1+e^xright] dx I = 48 int_0^fracpi2 x^2 (1 + cos 2x) dx ### Step 2: Evaluating the integrated components Integrating by parts gives: I = pi (2pi^2 - 12) Matching coefficients with the template: alpha = 2 and \beta = -12. (alpha + beta)^2 = (2 - 12)^2 = (-10)^2 = 100 ### Pattern Recognition Exponential denominators like 1+e^x in symmetric integral intervals are prime candidates for simplification using King's property. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integrals
Q68 jee_main_2025_04_april_evening Properties of Definite Integrals
Let f(x) + 2fleft(frac1xright) = x^2 + 5 and 2 mathrm g (mathrm x) - 3 mathrm g left(frac 12right) = mathrm x, mathrm x > 0. If alpha = int_ 1 ^ 2 f (x) d x, and beta = int_ 1 ^ 2 g (x) d x, then the value of 9alpha + beta is:
  • A. 1
  • B. 0
  • C. 10
  • D. 11

Solution

### Core Logic We have two functional equations to solve before integrating. Equation 1: f(x) + 2fleft(frac1xright) = x^2 + 5 Replace x with frac1x: fleft(frac1xright) + 2f(x) = frac1x^2 + 5 Multiplying this new equation by 2 and subtracting the original Equation 1 eliminates the fleft(frac1xright) term: 4f(x) + 2fleft(frac1xright) - left(f(x) + 2fleft(frac1xright)right) = 2left(frac1x^2 + 5right) - (x^2 + 5) 3f(x) = frac2x^2 - x^2 + 5 implies f(x) = frac23x^2 - fracx^23 + frac53 ### Step 1: Finding alpha Integrate f(x) from 1 to 2: alpha = int_1^2 left( frac23x^2 - fracx^23 + frac53 right) dx = left[ -frac23x - fracx^39 + frac5x3 right]_1^2 alpha = left( -frac13 - frac89 + frac103 right) - left( -frac23 - frac19 + frac53 right) = frac199 - frac89 = frac119 Thus, 9alpha = 11. ### Step 2: Solving for g(x) and finding beta We are given 2g(x) - 3gleft(frac12right) = x. Substitute x = frac12: 2gleft(frac12right) - 3gleft(frac12right) = frac12 implies -gleft(frac12right) = frac12 implies gleft(frac12right) = -frac12 Substitute this constant value back into the original equation: 2g(x) - 3left(-frac12right) = x implies 2g(x) + frac32 = x implies g(x) = fracx2 - frac34 Now find beta: beta = int_1^2 left( fracx2 - frac34 right) dx = left[ fracx^24 - frac3x4 right]_1^2 = left( 1 - frac32 right) - left( frac14 - frac34 right) = -frac12 - left(-frac12right) = 0 ### Step 3: Calculating 9alpha + beta Combining our values: 9alpha + beta = 11 + 0 = 11 ### Pattern Recognition Functional equations involving x to frac1x are easily solved by treating the swapped forms as a system of linear equations, allowing direct isolation of the underlying function. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Functional Equations
Q66 jee_main_2025_04_april_morning Properties of Definite Integrals
The value of int_-1^1fracleft(1 + sqrt|x| - xright)e^x + left(sqrt|x| - xright)e^-xe^x + e^-x \, mathrmdx is equal to
  • A. 3 - frac2sqrt23
  • B. 2 + frac2sqrt23
  • C. 1 - frac2sqrt23
  • D. 1 + frac2sqrt23

Solution

### Related Formula King's property of definite integrals: int_a^b f(x)mathrmdx = int_a^b f(a+b-x)mathrmdx ### Core Logic Let the given integral be I. Apply King's property by substituting x to -x: I = int_-1^1fracleft(1 + sqrt|x| + xright)e^-x + left(sqrt|x| + xright)e^xe^-x + e^x\,mathrmdx Add both integral expressions 2I = I + I: 2I = int_-1^1 frac(e^x + e^-x) + left(sqrt|x| - x + sqrt|x| + xright)(e^x + e^-x)e^x + e^-x\,mathrmdx 2I = int_-1^1 left(1 + sqrt|x| - x + sqrt|x| + xright)\,mathrmdx ### Step 1: Apply Symmetry Properties The integrand is completely even. Hence, convert intervals: 2I = 2int_0^1 left(1 + sqrt|x| - x + sqrt|x| + xright)\,mathrmdx For x in [0,1], |x| = x implies sqrt|x| - x = 0 and sqrt|x| + x = sqrt2x: I = int_0^1 (1 + sqrt2x)\,mathrmdx ### Step 2: Final Integration Execution I = left[ x + sqrt2 cdot fracx^3/23/2 right]_0^1 = left[ x + frac2sqrt23x^3/2 right]_0^1 I = 1 + frac2sqrt23 ### Pattern Recognition When functions involve combinations of exponential components (e^x, e^-x) over symmetric boundaries, adding the variable reflection eliminates exponential fractions instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration
Q52 jee_main_2025_24_jan_morning Properties of Definite Integrals
If I(m,n) = int_0^1 x^m-1 (1-x)^n-1 dx where m, n > 0, then I(9,14) + I(10,13) is :
  • A. I(9, 1)
  • B. I(19, 27)
  • C. I(1, 13)
  • D. I(9, 13)

Solution

### Related Formula The beta function integral format satisfies: I(m,n) = int_0^1 x^m-1 (1-x)^n-1 dx ### Core Logic Let's combine the terms of the requested sum directly by inserting their respective definitions: I(9,14) = int_0^1 x^9-1 (1-x)^14-1 dx = int_0^1 x^8 (1-x)^13 dx I(10,13) = int_0^1 x^10-1 (1-x)^13-1 dx = int_0^1 x^9 (1-x)^12 dx ### Step 1: Factoring out common algebraic terms Summing the two components: I(9,14) + I(10,13) = int_0^1 left[ x^8 (1-x)^13 + x^9 (1-x)^12 right] dx Factor out the common term x^8 (1-x)^12 inside the integrand: = int_0^1 x^8 (1-x)^12 left[ (1-x) + x right] dx = int_0^1 x^8 (1-x)^12 (1) dx = int_0^1 x^9-1 (1-x)^13-1 dx = I(9,13) ### Pattern Recognition When dealing with linear combinations of beta functions with shifting parameter indices, directly writing down the definite integral expression often results in immediate algebraic cancellation or simplification via basic factoring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

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