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Integral calculus

Integral calculus. XII STANDARD MATHEMATICS. PREPARED BY: R.RAJENDRAN. M.A., M . Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21. Evaluate:. Adding (1) and (2). 2I = 3 I = 3/2. Evaluate:. Adding (1) and (2). Evaluate:. Adding (1) and (2). Evaluate:.

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Integral calculus

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  1. Integral calculus XII STANDARD MATHEMATICS PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

  2. Evaluate: Adding (1) and (2) 2I = 3 I = 3/2

  3. Evaluate: Adding (1) and (2)

  4. Evaluate: Adding (1) and (2)

  5. Evaluate: Let u = x/4, then dx = 4du When x = 2, u = /2 When x = 0, u = 0

  6. Find the area of the circle whose radius is a. y x Equation of the circle whose center is origin and radius a units is x2 + y2 = a2. Since it is symmetrical about both the axes, The required area is 4times the area in the first quadrant. The required area =

  7. Find the area of the region bounded by the line y = 2x + 4, y = 1, y = 3 and y-axis y y =3 y =1 The required area lies to the left of y axis between y = 1 and y = 3 x y =2x+3 The required area = = 2sq.units

  8. Find the area of the region bounded by x2 = 36y, y-axis, y = 2 and y = 4. y y = 4 x2 = 36y The required area lies to the right of y-axis between y = 2 and y = 4 The required area = y = 2 x

  9. Find the volume of the solid that results when the ellipse (a > b > 0)is revolved about the minor axis. y The required volume is twice the volume obtained by revolving the area in the first quadrant about the minor axis (y-axis) between y = 0 and y = b x The required volume =

  10. Find the area between the curve y = x2 –x – 2, x-axis, and the lines x = – 2 , x = 4 y x=4 x=-2 x Equation of the curve is y = x2 – x – 2 When y = 0, x2 – x – 2 = 0 (x – 2)(x + 1) = 0 x = 2, – 1 The curve cuts x-axis at x = –1 and x = 2 The required area = A1 + A2 + A3 Where A1 is area above the x-axis between x = –2 and x = –1 A2 is area below the x-axis between x = –1 and x = 2 A3 is area above the x-axis between x = 2 and x = 4 The required area =

  11. y Find the area enclosed by the parabolas y2 = x and x2 = y x2 = y x x=1 Equation of the parabolas are y2 = x………(1) = f(x) x2 = y………(2) = g(x) Sub (2) in (1) (x2)2 = x x4 – x = 0 x(x3 – 1) = 0 x = 0, 1 If x = 1, y = 1 The point of intersection is (1, 1) The required area = area between the two curves from x = 0 to x = 1 Required area = y2 = x

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