## Vineet Loomba is teaching live on Unacademy Plus

MoST IMPORTANT QUESTIONS IN APPLICATIONS OF INTEGRALS JEE MAIN RANK BOOSTER COURSE PREPARED BY ER. VINEET LOOMBA ITIAN IIT-JEE MENTOR

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B. Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 * Many of my students are pursuing courses in IITs, NITs, BITs etc unacademy * Currently running my own Coaching Institute for JEE Main and Advanced Search vineet loomba unacademy" on GOOGLE Follow me to access all the courses free of cost.

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Follow me on Unacademy for Next live Special Class Strategy for JEE Main and Advanced V Limits, Continuity, Differentiability v Sets, Relations and Functions Indefinite Integration Trigonometry V Definite Integratiorn VApplications of Derivatives Complex Number Permutations Combinations Binomial Theorem V Logarithmic Functions V Sequences Series V Straight Lines Applications of Integrals Circles (Detailed Course) Probability Most Important Questions in IIT-IEE Mathematics Parabola (Detailed Course) Ellipse (Detailed Course) V Inverse Trigonometry Upcoming Courses next month: Mathematical Reasoning Coordinate Geometry Differential Equations

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) SPECIAL ANNOUNCEMENT Share useful links and advice with your learners... CRASH COURSE FOR JEE MAIN 2019 (JANUARY SESSION) Find courses & lessons 735 Courses IIT JEE 2,521 Courses Str Chemistry 797 Courses Details in Plus Tab wwEest HINDI Home My Library Plus Profile SUCCESS IN JEE MAIN ER. VINEET LoOMBA (IIT RooRKEE)

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) The area of the region bounded by the curves y = x2 and y = lxl is [IIT-JEE) Example: 3 3 Solution: (b) Required area SUCCESS IN EE MAIN ER. VINEET LOOMBA (IIT RooRKEe)

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example: The area of the region bounded by the curve y-2x-x and line y -x is [IIT-JEE] 2 3 (d 4 Solution: (d) The given curve is y 2x-x y=-(x2-2x + 1) + 1 y-1--(x-1)2, it represents a downward parabola with vertex (1,1) (O, 0) Its points of intersection with the line y = x are (0.0) and (1,1). Required area = shaded region SUCCESS IN EE MAIN ER. VINEET LOOMBA (IIT RooRKEe)

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The area bounded by the hyperbola xy a and the line x 2a is (A) Ba2 -a log (2+v3) (B) 243a -a log(2+3) (D) 23a2 -a log(2-3) (C) Ba-a log(2-3) SOLUTION (B) Shaded area 2x (Area of the portion above X- axis) The equation ofthe curve above x-axis is required area (A)= 2 x2-a2 dr Alternative Method: 2a Area (A) (2a-Va, + y2)dy -log | x + yB A 23 a-2 log (2a+Ba+a A-2V3 a--a. log (2+83). log a Solve it yourself to confirm. SUCCESS IN EE MAIN ER. VINEET LOOMBA (IIT RooRKEe)

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration is The area bounded by the parabola y x X-axis and the tangent to the parabola at (1, 1) 4 12 SOLUTION (D) The given curve isy-r. Equation oftangent at A = ( 1 , 1 ) is (x-1) dx [using : y-y, = m (x-x)] The point ofintersectionof(i) with X-axis is B (1/2,0). Shaded area-area (OACO)-area (ABC) 1/2 area =11 (1-1/2) B C > area=- 12 SUCCESS IN EE MAIN ER. VINEET LOOMBA (IIT RooRKEe)

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration: Find the area bounded by the curve x 2 v and y-axis. Solution: The required area (2-y-y')dy 9 = sq. units 2 SUCCESS IN EE MAIN ER. VINEET LOOMBA (IIT RooRKEe)

100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The area bounded by the curves : y-4a (x + a) and y-4b (b-x) is : (A) 5(a+bs4ab(B) a+bV4ab (C)bv4ab(D) None of these SOLUTION: (B) The two curves are: =4a (x + a) 2b 2a and y-4b (b-x) Solving 4a(r+a) and y 4b (b-x) simultaneously, we get the coordinates of A and B Replacing values of x from (ii) and (i), we get using property-8] 2 3b 4b y2 = 4b (b-x) y2 -4a (x +) >A- (b-a, V4ab) and B(b-a,-4ab) 4ab shaded area 4b4a SUCCESS IN EE MAIN ER. VINEET LOOMBA (IIT RooRKEe)