 # Integral calculus XII STANDARD MATHEMATICS. Evaluate: Adding (1) and (2) 2I = 3 I = 3/2.

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Integral calculus XII STANDARD MATHEMATICS

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

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

Find the area of the circle whose radius is a. Equation of the circle whose center is origin and radius a units is x 2 + y 2 = a 2. Since it is symmetrical about both the axes, The required area is 4times the area in the first quadrant. x y The required area =

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

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

Find the volume of the solid that results when the ellipse (a > b > 0)is revolved about the minor axis. 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 y The required volume =

Find the area between the curve y = x 2 –x – 2, x-axis, and the lines x = – 2, x = 4 Equation of the curve is y = x 2 – x – 2 When y = 0, x 2 – 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 = A 1 + A 2 + A 3 Where A 1 is area above the x-axis between x = –2 and x = –1 A 2 is area below the x-axis between x = –1 and x = 2 A 3 is area above the x-axis between x = 2 and x = 4 The required area = x y x=4 x=-2

Find the area enclosed by the parabolas y 2 = x and x 2 = y Equation of the parabolas are y 2 = x………(1) = f(x) x 2 = y………(2) = g(x) Sub (2) in (1) (x 2 ) 2 = x x 4 – x = 0 x(x 3 – 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 = x y y 2 = x x 2 = y x=1

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