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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 Application of Integral Solved Examples - Mathematics (Maths) for JEE Main & Advanced
| 1. What is the application of integrals in real-life scenarios? |  |
Ans. The application of integrals in real-life scenarios includes calculating areas under curves, finding volumes of solids, determining work done by a force, and analyzing population growth or decay.
| 2. How can integrals be used to calculate the area under a curve? | |
Ans. Integrals can be used to calculate the area under a curve by finding the definite integral of the function over a given interval. The result gives the area between the curve and the x-axis within that interval.
| 3. How are integrals helpful in determining the work done by a force? | |
Ans. Integrals are helpful in determining the work done by a force by integrating the force function with respect to distance. The result gives the total work done in moving an object along a specified path.
| 4. Can integrals be used to find the volume of irregular shapes or solids? | |
Ans. Yes, integrals can be used to find the volume of irregular shapes or solids by setting up appropriate integrals based on the dimensions of the object. By integrating over the desired region, the volume can be accurately calculated.
| 5. In what ways can integrals be utilized in analyzing population growth or decay? | |
Ans. Integrals can be utilized in analyzing population growth or decay by setting up differential equations that model the rate of change of population over time. By integrating these equations, one can predict future population trends and make informed decisions based on the data.
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