JEE Exam > JEE Notes > Mathematics (Maths) Main & Advanced > JEE Main Previous Year Questions (2026): Definite Integrals
JEE Main Previous Year Questions (2026): Definite Integrals
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 JEE Main Previous Year Questions (2025): Definite Integrals Q1: Let ?? : ( ?? , 8 ) ? ?? be a twice differentiable function. If for some ?? ? ?? , ? ? ?? ?? ?? ( ???? ) ?? ?? = ???? ( ?? ) , ?? ( ?? ) = ?? and ?? ( ???? ) = ?? ?? , then ???? - ?? ' ( ?? ???? ) is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 112 Solution: ? 0 1 ? f ( ?? x ) d ?? = af ( x ) ?? x = t d ?? = 1 x dt 1 ?? ? 0 ?? ? ?? ( ?? ) ???? = ???? ( ?? ) ? 0 x ? f ( t ) dt = ax f ( x ) f ( x ) = a ( xf ' ( x ) + f ( x ) ) ( 1 - ?? ) ?? ( ?? ) = ?? · ?? ?? ' ( ?? ) f ' ( x ) f ( x ) = ( 1 - a ) a 1 x l ln ? f ( x ) = 1 - a a l nx + c x = 1 , f ( 1 ) = 1 ? c = 0 x = 16 , f ( 16 ) = 1 8 1 8 = ( 16 ) 1 - ?? ?? ? - 3 = 4 - 4 ?? ?? ? a = 4 f ( x ) = x - 3 4 f ' ( x ) = - 3 4 x - 7 4 ? 16 - f ' ( 1 16 ) = 16 - ( - 3 4 ( 2 - 4 ) - 7 / 4 ) = 16 + 96 = 112 Page 2 JEE Main Previous Year Questions (2025): Definite Integrals Q1: Let ?? : ( ?? , 8 ) ? ?? be a twice differentiable function. If for some ?? ? ?? , ? ? ?? ?? ?? ( ???? ) ?? ?? = ???? ( ?? ) , ?? ( ?? ) = ?? and ?? ( ???? ) = ?? ?? , then ???? - ?? ' ( ?? ???? ) is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 112 Solution: ? 0 1 ? f ( ?? x ) d ?? = af ( x ) ?? x = t d ?? = 1 x dt 1 ?? ? 0 ?? ? ?? ( ?? ) ???? = ???? ( ?? ) ? 0 x ? f ( t ) dt = ax f ( x ) f ( x ) = a ( xf ' ( x ) + f ( x ) ) ( 1 - ?? ) ?? ( ?? ) = ?? · ?? ?? ' ( ?? ) f ' ( x ) f ( x ) = ( 1 - a ) a 1 x l ln ? f ( x ) = 1 - a a l nx + c x = 1 , f ( 1 ) = 1 ? c = 0 x = 16 , f ( 16 ) = 1 8 1 8 = ( 16 ) 1 - ?? ?? ? - 3 = 4 - 4 ?? ?? ? a = 4 f ( x ) = x - 3 4 f ' ( x ) = - 3 4 x - 7 4 ? 16 - f ' ( 1 16 ) = 16 - ( - 3 4 ( 2 - 4 ) - 7 / 4 ) = 16 + 96 = 112 Q2: If ?????? ?? ? ?? ? ( ? ?? ?? ? ( ?? ?? + ?? ) ?? ???? ) ?? ?? = ?? ?? ?? ( ?? ?? ) ?? ?? , then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 64 Solution: 1 8 form Now ?? = ?? ?? ? 0 1 ?? ( ( 3 ?? + 5 ) ?? + 1 3 ( ?? + 1 ) | 0 1 - 1 ) = ?? ?? ? 0 8 ?? + 1 - 5 ?? + 1 - 3 ?? - 3 3 ?? ( ?? + 1 ) = ?? 8ln ? 8 - 5ln ? 5 - 3 3 = ( 8 5 ) 2 / 3 ( 64 5 ) = ?? 5e ( 8 5 ) 2 / 3 On comparing ?? = 64 Q3: If ???? ? ?? ?? ?? ? [ ?? ???? ? ?? ?? - ?? ???? | ? + [ ???????? ? ?? ] ] ???? = ?? ?? + ?? , where [ · ] denotes the greatest integer function, then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: = 24 ? 0 ?? 48 ? - sin ? ( 4x - ?? 12 ) + ? ?? / 48 ?? / 4 ? s i n ? ( 4x - ?? 12 ) + ? 0 ?? 6 ? [ 0 ] ???? + ? ?? / 6 ?? / 4 ? [ 2s i n ? ?? ] ???? = 24 [ ( 1 - c os ? ?? 12 ) 4 - ( - c os ? ?? 12 - 1 ) 4 ] + ?? 4 - ?? 6 = 24 ( 1 2 + ?? 12 ) = 2 ?? + 12 ?? = 12 Q4: Let [.] denote the greatest integer function. If ? ?? ?? ?? ? [ ?? ?? ?? - ?? ] ???? = ?? - ???? ?? ?? ? ?? , then ?? ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 2nd April Morning Shift Ans: 8 Page 3 JEE Main Previous Year Questions (2025): Definite Integrals Q1: Let ?? : ( ?? , 8 ) ? ?? be a twice differentiable function. If for some ?? ? ?? , ? ? ?? ?? ?? ( ???? ) ?? ?? = ???? ( ?? ) , ?? ( ?? ) = ?? and ?? ( ???? ) = ?? ?? , then ???? - ?? ' ( ?? ???? ) is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 112 Solution: ? 0 1 ? f ( ?? x ) d ?? = af ( x ) ?? x = t d ?? = 1 x dt 1 ?? ? 0 ?? ? ?? ( ?? ) ???? = ???? ( ?? ) ? 0 x ? f ( t ) dt = ax f ( x ) f ( x ) = a ( xf ' ( x ) + f ( x ) ) ( 1 - ?? ) ?? ( ?? ) = ?? · ?? ?? ' ( ?? ) f ' ( x ) f ( x ) = ( 1 - a ) a 1 x l ln ? f ( x ) = 1 - a a l nx + c x = 1 , f ( 1 ) = 1 ? c = 0 x = 16 , f ( 16 ) = 1 8 1 8 = ( 16 ) 1 - ?? ?? ? - 3 = 4 - 4 ?? ?? ? a = 4 f ( x ) = x - 3 4 f ' ( x ) = - 3 4 x - 7 4 ? 16 - f ' ( 1 16 ) = 16 - ( - 3 4 ( 2 - 4 ) - 7 / 4 ) = 16 + 96 = 112 Q2: If ?????? ?? ? ?? ? ( ? ?? ?? ? ( ?? ?? + ?? ) ?? ???? ) ?? ?? = ?? ?? ?? ( ?? ?? ) ?? ?? , then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 64 Solution: 1 8 form Now ?? = ?? ?? ? 0 1 ?? ( ( 3 ?? + 5 ) ?? + 1 3 ( ?? + 1 ) | 0 1 - 1 ) = ?? ?? ? 0 8 ?? + 1 - 5 ?? + 1 - 3 ?? - 3 3 ?? ( ?? + 1 ) = ?? 8ln ? 8 - 5ln ? 5 - 3 3 = ( 8 5 ) 2 / 3 ( 64 5 ) = ?? 5e ( 8 5 ) 2 / 3 On comparing ?? = 64 Q3: If ???? ? ?? ?? ?? ? [ ?? ???? ? ?? ?? - ?? ???? | ? + [ ???????? ? ?? ] ] ???? = ?? ?? + ?? , where [ · ] denotes the greatest integer function, then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: = 24 ? 0 ?? 48 ? - sin ? ( 4x - ?? 12 ) + ? ?? / 48 ?? / 4 ? s i n ? ( 4x - ?? 12 ) + ? 0 ?? 6 ? [ 0 ] ???? + ? ?? / 6 ?? / 4 ? [ 2s i n ? ?? ] ???? = 24 [ ( 1 - c os ? ?? 12 ) 4 - ( - c os ? ?? 12 - 1 ) 4 ] + ?? 4 - ?? 6 = 24 ( 1 2 + ?? 12 ) = 2 ?? + 12 ?? = 12 Q4: Let [.] denote the greatest integer function. If ? ?? ?? ?? ? [ ?? ?? ?? - ?? ] ???? = ?? - ???? ?? ?? ? ?? , then ?? ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 2nd April Morning Shift Ans: 8 Solution: To solve this, we start by evaluating the integral: ?? = ? 0 ?? 3 ? [ 1 ?? ?? - 1 ] ???? The greatest integer function [ · ] returns the largest integer less than or equal to the input value. Here's how we can approach the problem: Determine the function inside the integral: 1 ?? ?? - 1 = ?? 1 - ?? . Identifying the intervals: When ?? 1 - ?? = 2, which simplifies to ?? = 1 - ln ? 2, we have [ ?? 1 - ?? ] = 2. When 1 = ?? 1 - ?? < 2, simplifying gives 1 - ln ? 2 < ?? = 1, and thus [ ?? 1 - ?? ] = 1. When 0 = ?? 1 - ?? < 1, which holds for ?? > 1, thus [ ?? 1 - ?? ] = 0 from ?? = 1 to ?? = ?? 3 . Evaluate the integral on these intervals: ? 0 1 - ln ? 2 ? 2 ???? = 2 ( 1 - ln ? 2 ) ? 1 - ln ? 2 1 ? 1 ???? = 1 - ( 1 - ln ? 2 ) = ln ? 2 ? 1 ?? 3 ? 0 ???? = 0 Combine these results: ?? = 2 ( 1 - ln ? 2 ) + ln ? 2 + 0 = 2 - ln ? 2 Thus, we are given that: ?? - ln ? 2 = 2 - ln ? 2 This implies that: ?? = 2 Therefore, ?? 3 = 2 3 = 8. Q5: Let for ?? ( ?? ) = ?? ?????? ?? ? ?? + ?? ?????? ?? ? ?? - ?? ?????? ?? ? ?? - ?? ?????? ?? ? ?? , ?? ?? = ? ?? ?? / ?? ? ?? ( ?? ) ?? ?? and ?? ?? = ? ?? ?? / ?? ? ???? ( ?? ) ?? ?? . Then ?? ?? ?? + ???? ?? ?? is equal to: JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 2 ?? B. 1 C. ?? D. 2 Ans: B Solution: Page 4 JEE Main Previous Year Questions (2025): Definite Integrals Q1: Let ?? : ( ?? , 8 ) ? ?? be a twice differentiable function. If for some ?? ? ?? , ? ? ?? ?? ?? ( ???? ) ?? ?? = ???? ( ?? ) , ?? ( ?? ) = ?? and ?? ( ???? ) = ?? ?? , then ???? - ?? ' ( ?? ???? ) is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 112 Solution: ? 0 1 ? f ( ?? x ) d ?? = af ( x ) ?? x = t d ?? = 1 x dt 1 ?? ? 0 ?? ? ?? ( ?? ) ???? = ???? ( ?? ) ? 0 x ? f ( t ) dt = ax f ( x ) f ( x ) = a ( xf ' ( x ) + f ( x ) ) ( 1 - ?? ) ?? ( ?? ) = ?? · ?? ?? ' ( ?? ) f ' ( x ) f ( x ) = ( 1 - a ) a 1 x l ln ? f ( x ) = 1 - a a l nx + c x = 1 , f ( 1 ) = 1 ? c = 0 x = 16 , f ( 16 ) = 1 8 1 8 = ( 16 ) 1 - ?? ?? ? - 3 = 4 - 4 ?? ?? ? a = 4 f ( x ) = x - 3 4 f ' ( x ) = - 3 4 x - 7 4 ? 16 - f ' ( 1 16 ) = 16 - ( - 3 4 ( 2 - 4 ) - 7 / 4 ) = 16 + 96 = 112 Q2: If ?????? ?? ? ?? ? ( ? ?? ?? ? ( ?? ?? + ?? ) ?? ???? ) ?? ?? = ?? ?? ?? ( ?? ?? ) ?? ?? , then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 64 Solution: 1 8 form Now ?? = ?? ?? ? 0 1 ?? ( ( 3 ?? + 5 ) ?? + 1 3 ( ?? + 1 ) | 0 1 - 1 ) = ?? ?? ? 0 8 ?? + 1 - 5 ?? + 1 - 3 ?? - 3 3 ?? ( ?? + 1 ) = ?? 8ln ? 8 - 5ln ? 5 - 3 3 = ( 8 5 ) 2 / 3 ( 64 5 ) = ?? 5e ( 8 5 ) 2 / 3 On comparing ?? = 64 Q3: If ???? ? ?? ?? ?? ? [ ?? ???? ? ?? ?? - ?? ???? | ? + [ ???????? ? ?? ] ] ???? = ?? ?? + ?? , where [ · ] denotes the greatest integer function, then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: = 24 ? 0 ?? 48 ? - sin ? ( 4x - ?? 12 ) + ? ?? / 48 ?? / 4 ? s i n ? ( 4x - ?? 12 ) + ? 0 ?? 6 ? [ 0 ] ???? + ? ?? / 6 ?? / 4 ? [ 2s i n ? ?? ] ???? = 24 [ ( 1 - c os ? ?? 12 ) 4 - ( - c os ? ?? 12 - 1 ) 4 ] + ?? 4 - ?? 6 = 24 ( 1 2 + ?? 12 ) = 2 ?? + 12 ?? = 12 Q4: Let [.] denote the greatest integer function. If ? ?? ?? ?? ? [ ?? ?? ?? - ?? ] ???? = ?? - ???? ?? ?? ? ?? , then ?? ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 2nd April Morning Shift Ans: 8 Solution: To solve this, we start by evaluating the integral: ?? = ? 0 ?? 3 ? [ 1 ?? ?? - 1 ] ???? The greatest integer function [ · ] returns the largest integer less than or equal to the input value. Here's how we can approach the problem: Determine the function inside the integral: 1 ?? ?? - 1 = ?? 1 - ?? . Identifying the intervals: When ?? 1 - ?? = 2, which simplifies to ?? = 1 - ln ? 2, we have [ ?? 1 - ?? ] = 2. When 1 = ?? 1 - ?? < 2, simplifying gives 1 - ln ? 2 < ?? = 1, and thus [ ?? 1 - ?? ] = 1. When 0 = ?? 1 - ?? < 1, which holds for ?? > 1, thus [ ?? 1 - ?? ] = 0 from ?? = 1 to ?? = ?? 3 . Evaluate the integral on these intervals: ? 0 1 - ln ? 2 ? 2 ???? = 2 ( 1 - ln ? 2 ) ? 1 - ln ? 2 1 ? 1 ???? = 1 - ( 1 - ln ? 2 ) = ln ? 2 ? 1 ?? 3 ? 0 ???? = 0 Combine these results: ?? = 2 ( 1 - ln ? 2 ) + ln ? 2 + 0 = 2 - ln ? 2 Thus, we are given that: ?? - ln ? 2 = 2 - ln ? 2 This implies that: ?? = 2 Therefore, ?? 3 = 2 3 = 8. Q5: Let for ?? ( ?? ) = ?? ?????? ?? ? ?? + ?? ?????? ?? ? ?? - ?? ?????? ?? ? ?? - ?? ?????? ?? ? ?? , ?? ?? = ? ?? ?? / ?? ? ?? ( ?? ) ?? ?? and ?? ?? = ? ?? ?? / ?? ? ???? ( ?? ) ?? ?? . Then ?? ?? ?? + ???? ?? ?? is equal to: JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 2 ?? B. 1 C. ?? D. 2 Ans: B Solution: ?? ( ?? ) ? = 7 tan 8 ? ?? + 7 tan 6 ? ?? - 3 tan 4 ? ?? - 3 tan 2 ? ?? ? = 7 tan 6 ? ?? ( 1 + tan 2 ? ?? ) - 3 tan 2 ? ?? ( 1 + tan 2 ? ?? ) ? = ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) ( 1 + tan 2 ? ?? ) ? = ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) s ec 2 ? ?? ?? 1 = ? ? ? ?? 4 0 ? ?? ( ?? ) ???? = ? ? ?? 4 0 ? ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) s ec 2 ? ???? ?? ? = ( 7 tan 7 ? ?? 7 - 3 tan 3 ? ?? 3 ) | 0 ?? 4 = 1 - 1 = 0 ?? 2 = ? ? ? ?? 4 0 ? ???? ( ?? ) ???? = ? ? ?? 4 0 ? ?? ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) s ec 2 ? ???? ?? ? = ?? ( tan 7 ? ?? - tan 3 ? ?? ) | 0 ?? 4 - ? ? ?? 4 0 ? 1 · ( tan 7 ? ?? - tan 3 ? ?? ) ???? ? = 0 - ? ? ?? 4 0 ? tan 3 ? ?? ( tan 2 ? ?? - 1 ) ( tan 2 ? ?? + 1 ) ???? = ? 0 ?? 4 ? ( tan 3 ? ?? - tan 5 ? ?? ) s e c 2 ? ???? ?? = tan 4 ? ?? 4 - tan 6 ? ?? 6 | 0 ?? 4 = 1 12 Hence 7 ?? 1 + 12 ?? 2 = 1 Q6: The value of ? ?? ?? ?? ?? ? ?? ?? ( ?? ( ( ?? ???? ?? ? ?? ) ?? + ?? ) - ?? ?? ( ( ?? ???? ?? ? ?? ) ?? + ?? ) - ?? + ?? ( ( ?? - ?? ???? ?? ? ?? ) ?? + ?? ) - ?? ) ???? is Options: A. 1 B. log ?? ? 2 C. ?? 2 D. 2 Ans: A Solution: Let ln ? x = t ? dx x = dt ?? = ? 2 4 ? ?? 1 1 + ?? 2 ?? 1 1 + ?? 2 + ?? 1 1 + ( 6 - ?? ) 2 ???? Page 5 JEE Main Previous Year Questions (2025): Definite Integrals Q1: Let ?? : ( ?? , 8 ) ? ?? be a twice differentiable function. If for some ?? ? ?? , ? ? ?? ?? ?? ( ???? ) ?? ?? = ???? ( ?? ) , ?? ( ?? ) = ?? and ?? ( ???? ) = ?? ?? , then ???? - ?? ' ( ?? ???? ) is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 112 Solution: ? 0 1 ? f ( ?? x ) d ?? = af ( x ) ?? x = t d ?? = 1 x dt 1 ?? ? 0 ?? ? ?? ( ?? ) ???? = ???? ( ?? ) ? 0 x ? f ( t ) dt = ax f ( x ) f ( x ) = a ( xf ' ( x ) + f ( x ) ) ( 1 - ?? ) ?? ( ?? ) = ?? · ?? ?? ' ( ?? ) f ' ( x ) f ( x ) = ( 1 - a ) a 1 x l ln ? f ( x ) = 1 - a a l nx + c x = 1 , f ( 1 ) = 1 ? c = 0 x = 16 , f ( 16 ) = 1 8 1 8 = ( 16 ) 1 - ?? ?? ? - 3 = 4 - 4 ?? ?? ? a = 4 f ( x ) = x - 3 4 f ' ( x ) = - 3 4 x - 7 4 ? 16 - f ' ( 1 16 ) = 16 - ( - 3 4 ( 2 - 4 ) - 7 / 4 ) = 16 + 96 = 112 Q2: If ?????? ?? ? ?? ? ( ? ?? ?? ? ( ?? ?? + ?? ) ?? ???? ) ?? ?? = ?? ?? ?? ( ?? ?? ) ?? ?? , then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 64 Solution: 1 8 form Now ?? = ?? ?? ? 0 1 ?? ( ( 3 ?? + 5 ) ?? + 1 3 ( ?? + 1 ) | 0 1 - 1 ) = ?? ?? ? 0 8 ?? + 1 - 5 ?? + 1 - 3 ?? - 3 3 ?? ( ?? + 1 ) = ?? 8ln ? 8 - 5ln ? 5 - 3 3 = ( 8 5 ) 2 / 3 ( 64 5 ) = ?? 5e ( 8 5 ) 2 / 3 On comparing ?? = 64 Q3: If ???? ? ?? ?? ?? ? [ ?? ???? ? ?? ?? - ?? ???? | ? + [ ???????? ? ?? ] ] ???? = ?? ?? + ?? , where [ · ] denotes the greatest integer function, then ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: = 24 ? 0 ?? 48 ? - sin ? ( 4x - ?? 12 ) + ? ?? / 48 ?? / 4 ? s i n ? ( 4x - ?? 12 ) + ? 0 ?? 6 ? [ 0 ] ???? + ? ?? / 6 ?? / 4 ? [ 2s i n ? ?? ] ???? = 24 [ ( 1 - c os ? ?? 12 ) 4 - ( - c os ? ?? 12 - 1 ) 4 ] + ?? 4 - ?? 6 = 24 ( 1 2 + ?? 12 ) = 2 ?? + 12 ?? = 12 Q4: Let [.] denote the greatest integer function. If ? ?? ?? ?? ? [ ?? ?? ?? - ?? ] ???? = ?? - ???? ?? ?? ? ?? , then ?? ?? is equal to _ _ _ _ . JEE Main 2025 (Online) 2nd April Morning Shift Ans: 8 Solution: To solve this, we start by evaluating the integral: ?? = ? 0 ?? 3 ? [ 1 ?? ?? - 1 ] ???? The greatest integer function [ · ] returns the largest integer less than or equal to the input value. Here's how we can approach the problem: Determine the function inside the integral: 1 ?? ?? - 1 = ?? 1 - ?? . Identifying the intervals: When ?? 1 - ?? = 2, which simplifies to ?? = 1 - ln ? 2, we have [ ?? 1 - ?? ] = 2. When 1 = ?? 1 - ?? < 2, simplifying gives 1 - ln ? 2 < ?? = 1, and thus [ ?? 1 - ?? ] = 1. When 0 = ?? 1 - ?? < 1, which holds for ?? > 1, thus [ ?? 1 - ?? ] = 0 from ?? = 1 to ?? = ?? 3 . Evaluate the integral on these intervals: ? 0 1 - ln ? 2 ? 2 ???? = 2 ( 1 - ln ? 2 ) ? 1 - ln ? 2 1 ? 1 ???? = 1 - ( 1 - ln ? 2 ) = ln ? 2 ? 1 ?? 3 ? 0 ???? = 0 Combine these results: ?? = 2 ( 1 - ln ? 2 ) + ln ? 2 + 0 = 2 - ln ? 2 Thus, we are given that: ?? - ln ? 2 = 2 - ln ? 2 This implies that: ?? = 2 Therefore, ?? 3 = 2 3 = 8. Q5: Let for ?? ( ?? ) = ?? ?????? ?? ? ?? + ?? ?????? ?? ? ?? - ?? ?????? ?? ? ?? - ?? ?????? ?? ? ?? , ?? ?? = ? ?? ?? / ?? ? ?? ( ?? ) ?? ?? and ?? ?? = ? ?? ?? / ?? ? ???? ( ?? ) ?? ?? . Then ?? ?? ?? + ???? ?? ?? is equal to: JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 2 ?? B. 1 C. ?? D. 2 Ans: B Solution: ?? ( ?? ) ? = 7 tan 8 ? ?? + 7 tan 6 ? ?? - 3 tan 4 ? ?? - 3 tan 2 ? ?? ? = 7 tan 6 ? ?? ( 1 + tan 2 ? ?? ) - 3 tan 2 ? ?? ( 1 + tan 2 ? ?? ) ? = ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) ( 1 + tan 2 ? ?? ) ? = ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) s ec 2 ? ?? ?? 1 = ? ? ? ?? 4 0 ? ?? ( ?? ) ???? = ? ? ?? 4 0 ? ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) s ec 2 ? ???? ?? ? = ( 7 tan 7 ? ?? 7 - 3 tan 3 ? ?? 3 ) | 0 ?? 4 = 1 - 1 = 0 ?? 2 = ? ? ? ?? 4 0 ? ???? ( ?? ) ???? = ? ? ?? 4 0 ? ?? ( 7 tan 6 ? ?? - 3 tan 2 ? ?? ) s ec 2 ? ???? ?? ? = ?? ( tan 7 ? ?? - tan 3 ? ?? ) | 0 ?? 4 - ? ? ?? 4 0 ? 1 · ( tan 7 ? ?? - tan 3 ? ?? ) ???? ? = 0 - ? ? ?? 4 0 ? tan 3 ? ?? ( tan 2 ? ?? - 1 ) ( tan 2 ? ?? + 1 ) ???? = ? 0 ?? 4 ? ( tan 3 ? ?? - tan 5 ? ?? ) s e c 2 ? ???? ?? = tan 4 ? ?? 4 - tan 6 ? ?? 6 | 0 ?? 4 = 1 12 Hence 7 ?? 1 + 12 ?? 2 = 1 Q6: The value of ? ?? ?? ?? ?? ? ?? ?? ( ?? ( ( ?? ???? ?? ? ?? ) ?? + ?? ) - ?? ?? ( ( ?? ???? ?? ? ?? ) ?? + ?? ) - ?? + ?? ( ( ?? - ?? ???? ?? ? ?? ) ?? + ?? ) - ?? ) ???? is Options: A. 1 B. log ?? ? 2 C. ?? 2 D. 2 Ans: A Solution: Let ln ? x = t ? dx x = dt ?? = ? 2 4 ? ?? 1 1 + ?? 2 ?? 1 1 + ?? 2 + ?? 1 1 + ( 6 - ?? ) 2 ???? ?? = ? 2 4 ? 1 ?? 1 + ( 6 - ?? ) 2 1 ?? 1 + ( 6 - ?? ) 2 + ?? 1 1 + ?? 2 ???? 2 ?? = ? 2 4 ????? = ( ?? ) 2 4 = 4 - 2 = 2 ?? = 1 Q7: If ?? = ? ?? ?? ?? ? ?? ?? ?? ?? ?? ? ?? ?? ?? ?? ?? ?? ? ?? + ?? ?? ?? ?? ?? ? ?? ?? ?? , then ? ?? ?? ?? ? ?? ?? ?? ?? ? ?? ?? ?? ?? ? ?? ?? ?? ?? ?? ? ?? + ?? ?? ?? ?? ? ?? ?? ?? equals : JEE Main 2025 (Online) 23rd January Evening Shift Options: A. ?? 2 12 B. ?? 2 4 C. ?? 2 16 D. ?? 2 8 Ans: C Solution: For I Apply king (P - 5) and add 2 ?? = ? 0 ?? / 2 ????? = ?? 2 ? ?? = ?? 4 ?? 2 = ? 0 ?? / 2 ? ?? sin ? ?? c os ? ?? sin 4 ? ?? + c os 4 ? ?? ???? Apply king and add I 2 = ?? 4 ? 0 ?? / 2 ? tan ? x s ec 2 x dx tan 4 ? x + 1 put tan 2 ? x = t ?? 8 ? 0 8 ? dt t 2 + 1 = ?? 8 · ?? 2 = ?? 2 16 Q8: If ?? ( ?? , ?? ) = ? ?? ?? ? ?? ?? - ?? ( ?? - ?? ) ?? - ?? ???? , ?? , ?? > ?? , then ?? ( ?? , ???? ) + ?? ( ???? , ???? ) is JEE Main 2025 (Online) 24th January Morning Shift Options:Read More
FAQs on JEE Main Previous Year Questions (2026): Definite Integrals
| 1. What are definite integrals and why are they important in calculus? |  |
Ans.Definite integrals represent the signed area under a curve between two points on the x-axis. They are crucial for calculating quantities such as total distance, area, and accumulated values, providing a fundamental connection between algebra and geometry. Definite integrals are essential for solving problems related to motion, physics, and engineering.
| 2. How do you evaluate a definite integral using the Fundamental Theorem of Calculus? | |
Ans.To evaluate a definite integral using the Fundamental Theorem of Calculus, you first find the antiderivative (indefinite integral) of the function. Then, you compute the difference between the values of this antiderivative at the upper and lower limits of integration. Mathematically, if F is the antiderivative of f, then the definite integral from a to b can be calculated as F(b) - F(a).
| 3. What are some common techniques for solving definite integrals? | |
Ans.Common techniques for solving definite integrals include: 1. Substitution: Changing variables to simplify the integral. 2. Integration by parts: Using the product rule in reverse to integrate the product of two functions. 3. Partial fractions: Breaking down rational functions into simpler fractions that can be integrated individually. 4. Numerical integration: Approximating the value of the integral using methods like the trapezoidal rule or Simpson's rule.
| 4. Can definite integrals yield negative values? | |
Ans.Yes, definite integrals can yield negative values. This occurs when the curve lies below the x-axis between the limits of integration. In such cases, the integral represents the signed area, which is considered negative when computed.
| 5. What is the significance of the limits of integration in definite integrals? | |
Ans.The limits of integration in definite integrals define the interval over which the function is being analyzed. They determine the starting and ending points on the x-axis for calculating the area under the curve. Changing these limits can alter the value of the integral, reflecting different areas or accumulations related to the function being integrated.
About this Document
4.86/5 Rating
Apr 25, 2026 Last updated
Related Exams
Document Description: JEE Main Previous Year Questions (2026): Definite Integrals for JEE 2026 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The notes and questions for JEE Main Previous Year Questions (2026): Definite Integrals have been prepared according to the JEE exam syllabus. Information about JEE Main Previous Year Questions (2026): Definite Integrals covers topics like and JEE Main Previous Year Questions (2026): Definite Integrals Example, for JEE 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main Previous Year Questions (2026): Definite Integrals.
Introduction of JEE Main Previous Year Questions (2026): Definite Integrals in English is available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & JEE Main Previous Year Questions (2026): Definite Integrals in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: JEE Main Previous Year Questions (2026): Definite Integrals
Description
JEE Main Previous Year Questions (2026): Definite Integrals of Maths is important for the exam. Get access to all the questions with solutions asked in past years of JEE exam. Download it from EduRev.
Information about JEE Main Previous Year Questions (2026): Definite Integrals
In this doc you can find the meaning of JEE Main Previous Year Questions (2026): Definite Integrals defined & explained in the simplest way possible. Besides explaining types of JEE Main Previous Year Questions (2026): Definite Integrals theory, EduRev gives you an ample number of questions to practice JEE Main Previous Year Questions (2026): Definite Integrals tests, examples and also practice JEE tests
Related Searches
mock tests for examination, MCQs, Objective type Questions, past year papers, Semester Notes, JEE Main Previous Year Questions (2026): Definite Integrals, Viva Questions, shortcuts and tricks, Extra Questions, JEE Main Previous Year Questions (2026): Definite Integrals, Summary, pdf , Free, practice quizzes, Exam, study material, Sample Paper, video lectures, Important questions, Previous Year Questions with Solutions, ppt, JEE Main Previous Year Questions (2026): Definite Integrals;