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Page 1 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown Page 2 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure Page 3 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure ? Required area = 4 ? ? ?? 0 ? v?? 2 - ?? 2 ???? = 4 [ ?? 2 v?? 2 - ?? 2 + ?? 2 2 sin -1 ?? ?? ] 0 ?? = 4 [ ?? 2 × ?? 2 2 ] = ?? ?? 2 Q6: Find the area bounded by the parabola ?? ?? = ?? ?? and its latus rectum [ (a) ?? ?? (b) ?? ?? (c) ???? ?? (d) None of these Ans: (a) Since the curve is symmetrical about ?? -axis, therefore the required area = 2 ? ? 1 0 ??????? = 2 ? ? 1 0 ?v 4?? ???? = 4 · 2 3 [?? 3 2 ] 0 1 = 8 3 Page 4 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure ? Required area = 4 ? ? ?? 0 ? v?? 2 - ?? 2 ???? = 4 [ ?? 2 v?? 2 - ?? 2 + ?? 2 2 sin -1 ?? ?? ] 0 ?? = 4 [ ?? 2 × ?? 2 2 ] = ?? ?? 2 Q6: Find the area bounded by the parabola ?? ?? = ?? ?? and its latus rectum [ (a) ?? ?? (b) ?? ?? (c) ???? ?? (d) None of these Ans: (a) Since the curve is symmetrical about ?? -axis, therefore the required area = 2 ? ? 1 0 ??????? = 2 ? ? 1 0 ?v 4?? ???? = 4 · 2 3 [?? 3 2 ] 0 1 = 8 3 Q7: The area bounded by the curve ?? ?? = ?? ?? and ?? ?? = ?? ?? is (a) ???? ?? sq. units (b) ?? ???? sq. units (c) ???? ?? sq. units (d) ?? ???? sq. units Ans: (a) Required area = ? 0 4 ?( ???????? - ???????? ) Region = ? 0 4 ?(v 4?? - ?? 2 4 )???? = 16 3 square unit. Trick : From Important Tips' the area of the region bounded by ?? 2 = 4???? and ?? 2 = 4???? is 16???? 3 square unit. Here ?? 2 = 4?? and ?? 2 = 4?? , so ?? = 1 and ?? = 1 Required area = 16 3 ( 1) ( 1) = 16 3 square unit. Q8: The area of the bounded region by the curve ?? = ?????? ?? , the ?? -axis and the line ?? = ?? and ?? = ?? is (a) 4 (b) 2 (c) ?? (d) None of these Ans: (b) Required area = ? ? ?? 0 ?sin ?????? = 2 ? ? ?? /2 0 ?sin ?????? = 2[-cos ?? ] 0 ?? /2 = 2[( -cos ?? /2)- ( -cos 0) ] = 2( 1) = 2 square unit. Page 5 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure ? Required area = 4 ? ? ?? 0 ? v?? 2 - ?? 2 ???? = 4 [ ?? 2 v?? 2 - ?? 2 + ?? 2 2 sin -1 ?? ?? ] 0 ?? = 4 [ ?? 2 × ?? 2 2 ] = ?? ?? 2 Q6: Find the area bounded by the parabola ?? ?? = ?? ?? and its latus rectum [ (a) ?? ?? (b) ?? ?? (c) ???? ?? (d) None of these Ans: (a) Since the curve is symmetrical about ?? -axis, therefore the required area = 2 ? ? 1 0 ??????? = 2 ? ? 1 0 ?v 4?? ???? = 4 · 2 3 [?? 3 2 ] 0 1 = 8 3 Q7: The area bounded by the curve ?? ?? = ?? ?? and ?? ?? = ?? ?? is (a) ???? ?? sq. units (b) ?? ???? sq. units (c) ???? ?? sq. units (d) ?? ???? sq. units Ans: (a) Required area = ? 0 4 ?( ???????? - ???????? ) Region = ? 0 4 ?(v 4?? - ?? 2 4 )???? = 16 3 square unit. Trick : From Important Tips' the area of the region bounded by ?? 2 = 4???? and ?? 2 = 4???? is 16???? 3 square unit. Here ?? 2 = 4?? and ?? 2 = 4?? , so ?? = 1 and ?? = 1 Required area = 16 3 ( 1) ( 1) = 16 3 square unit. Q8: The area of the bounded region by the curve ?? = ?????? ?? , the ?? -axis and the line ?? = ?? and ?? = ?? is (a) 4 (b) 2 (c) ?? (d) None of these Ans: (b) Required area = ? ? ?? 0 ?sin ?????? = 2 ? ? ?? /2 0 ?sin ?????? = 2[-cos ?? ] 0 ?? /2 = 2[( -cos ?? /2)- ( -cos 0) ] = 2( 1) = 2 square unit. Trick : For the curve ?? = sin ?? or cos ?? , the area of ? 0 ?? /2 ?sin ?????? = 1, ? 0 ?? ?sin ?????? = 2, ? 0 3?? /2 ?sin ?????? = 3, ? 0 2?? ?sin ?????? = 4 and so on. Q9: The area enclosed by the parabola ?? ?? = ?? ?? and the line ?? = ?? ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Solve the equation ?? 2 = 8?? and the line ?? = 2?? , we get the point of intersection. Then find the required area bounded by this region. It is 4 3 . Trick : Required area = 8( 2) 2 3( 2) 3 = 32 24 = 4 3 [? Area bounded by ?? 2 = 4???? and ?? = ???? is 8?? 2 3?? 3 . Here ?? = 2, ?? = 2] Q10: If the area bounded by ?? = ?? ?? ?? and ?? = ?? ?? ?? , ?? > ?? , is 1 , then ?? = (a) 1 (b) ?? v ?? (c) ?? ?? (d) - ?? v ?? Ans: (b) The ?? coordinate of A is 1 ?? According to the given conditionRead More
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FAQs on Solved Example: Application of Integral - Mathematics (Maths) for JEE Main & Advanced
| 1. What is the application of integrals in JEE? |  |
Ans. In JEE, the application of integrals is used to solve problems related to finding areas, volumes, and even calculating work done in physics problems. It is an essential concept in calculus that helps in solving real-life problems.
| 2. How can integrals be used to find the area under a curve in JEE? | |
Ans. Integrals can be used to find the area under a curve by setting up definite integrals with appropriate limits of integration. By integrating the function over the interval, the area under the curve can be calculated accurately.
| 3. Can integrals be used to calculate the volume of a solid in JEE? | |
Ans. Yes, integrals can be used to calculate the volume of a solid by setting up triple integrals for three-dimensional shapes or using double integrals for two-dimensional shapes. By integrating the appropriate function, the volume of the solid can be determined.
| 4. How does the concept of integrals help in solving optimization problems in JEE? | |
Ans. The concept of integrals helps in solving optimization problems in JEE by setting up an equation to be optimized and using calculus techniques to find the maximum or minimum value. Integrals play a crucial role in determining the optimal solution in such problems.
| 5. What are some practical applications of integrals in JEE beyond calculus problems? | |
Ans. Some practical applications of integrals in JEE beyond calculus problems include using integrals in physics to calculate work done, in economics to find total revenue or profit, and in engineering to analyze stress distribution in structures. Integrals have diverse applications in various fields beyond mathematics.
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