source: ncert
topic: calculus
class_level: class_12
chapter: ch8
difficulty: advanced
Class 12 | Ch8: Application of Integrals AREA BETWEEN CURVE AND X-AXIS: Area = integral[a to b] |f(x)| dx. If f(x)>=0 on [a,b]: Area=integral[a to b] f(x) dx. If f(x)<=0 on [a,b]: Area=|integral[a to b] f(x) dx|=-integral[a to b] f(x) dx. If f changes sign: split at zeros, add absolute values of each part. AREA BETWEEN TWO CURVES: If f(x)>=g(x) on [a,b]: Area=integral[a to b] [f(x)-g(x)] dx. Always integrate UPPER curve minus LOWER curve. AREA USING Y-AXIS: Area=integral[c to d] |g(y)| dy where x=g(y) is curve expressed in terms of y. STANDARD AREAS: Circle x^2+y^2=r^2: total area=pir^2. Using integration: 4×integral[0 to r] sqrt(r^2-x^2) dx. Ellipse x^2/a^2+y^2/b^2=1: area=piab. Parabola y^2=4ax and line x=h: area=(4/3)hsqrt(ah) [to vertex from x=0 to x=h side]. SOLVED EXAMPLES: Example 1: Area bounded by y=x^2 and y=x+2. Intersection: x^2=x+2 → x^2-x-2=0 → (x-2)(x+1)=0 → x=-1,2. Area=integral[-1 to 2] (x+2-x^2) dx=[x^2/2+2x-x^3/3] from -1 to 2. =(2+4-8/3)-(-1/2-2+1/3... wait: At x=2: 4/2+4-8/3=2+4-8/3=6-8/3=10/3. At x=-1: 1/2-2+1/3=-7/6. Area=10/3-(-7/6)=10/3+7/6=20/6+7/6=27/6=9/2 sq units. Example 2: Area of circle x^2+y^2=16. Area=4×integral[0 to 4] sqrt(16-x^2) dx=4×[x/2sqrt(16-x^2)+8sin^(-1)(x/4)] from 0 to 4. =4×[0+8pi/2-0]=4×4pi=16pi sq units. (Confirms pir^2=pi*16.) Example 3: Area between y=x^2 and y=sqrt(x). Intersection: x^2=sqrt(x) → x^4=x → x(x^3-1)=0 → x=0,1. sqrt(x)>=x^2 on [0,1]. Area=integral0 to 1dx=[2x^(3/2)/3-x^3/3] from 0 to 1=2/3-1/3=1/3. COMMON MISTAKES: Always integrate UPPER minus LOWER (not absolute value of each separately). Find intersection points by setting curves equal. Area is always POSITIVE (use absolute values if needed).