Question 1

The area of the region \ (x, y) : x^2 - 8x y -x\ is :

Accepted Answer

The given region is bounded by the parabola y = x^2 - 8x and the line y = -x. To find the points of intersection, equate the two equations: x^2 - 8x = -x x^2 - 7x = 0 x(x - 7) = 0 The points of intersection are x = 0 and x = 7. In the interval [0, 7], the line y = -x lies above the parabola y = x^2 - 8x. The required area A is given by: A = _ 0 ^ 7 (-x - (x^2 - 8x)) dx A = _ 0 ^ 7 (7x - x^2) dx Evaluating the integral: A = [ 7x^2 2 - x^3 3 ]_ 0 ^ 7 A = 7(49) 2 - 343 3 A = 343 2 - 343 3 A = 343 (

Question 2

The area of the region \ (x, y) : 0 y 6 - x, y^2 4x - 3, x 0\ is:

Accepted Answer

The given region is defined by the inequalities: x 0 0 y 6 - x y 0 and x 6 - y y^2 4x - 3 x y^2 + 3 4 From these inequalities, for a given y 0, the value of x ranges from 0 to (6 - y, y^2 + 3 4 ). To find the point where the two bounding curves intersect, we equate them: 6 - y = y^2 + 3 4 24 - 4y = y^2 + 3 y^2 + 4y - 21 = 0 (y + 7)(y - 3) = 0 Since y 0, the intersection occurs at y = 3. For 0 y 3, the right boundary is the parabola x = y^2 + 3 4 . For 3 y 6, the right boundary is the line x = 6

Question 3

The area of the region R = \ (x, y): xy 27, 1 y x^2\ is equal to:

Accepted Answer

The given region is bounded by the curves y = 1, y = x^2, and xy = 27 in the first quadrant. Let us find the points of intersection of these curves: Intersection of y = 1 and y = x^2 gives x = 1. Intersection of y = x^2 and xy = 27 gives x(x^2) = 27 x^3 = 27 x = 3, so y = 9. Intersection of y = 1 and xy = 27 gives x = 27. We can find the area by integrating with respect to y from y = 1 to y = 9. For a given y, the value of x ranges from the parabola x = y to the hyperbola x = 27 y . The area A i

Question 4

The area of the region bounded by the curves x+3y^2=0 and x+4y^2=1 is equal to:

Accepted Answer

The equations of the given curves are x = -3y^2 and x = 1 - 4y^2. To find the points of intersection, equate the values of x: -3y^2 = 1 - 4y^2 y^2 = 1 y = 1 The region is bounded between y = -1 and y = 1. In this interval, 1 - 4y^2 -3y^2. The required area A is given by: A = _ -1 ^ 1 ( (1 - 4y^2) - (-3y^2) ) dy A = _ -1 ^ 1 (1 - y^2) dy Since 1 - y^2 is an even function: A = 2 _ 0 ^ 1 (1 - y^2) dy A = 2 [ y - y^3 3 ]_ 0 ^ 1 A = 2 ( 1 - 1 3 ) = 2 ( 2 3 ) = 4 3 Answer: 4 3

Question 5

The area of the region \ (x, y): y - |x|, y |x x|, y 0\ is:

Accepted Answer

The given region is defined by the inequalities y - |x|, y |x x|, and y 0. Since replacing x with -x leaves the inequalities unchanged, the region is symmetric with respect to the y-axis. We can find the area of the region in the first quadrant (x 0) and multiply it by 2. For x 0, the inequalities become: y - x y x x y 0 The condition y 0 and y - x restricts x to the interval [0, ]. In this interval, x 0, so |x x| = x x. To find the intersection of the curves y = - x and y = x x, we equate them:

Question 6

If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y = 24 is A, then 3(A + 6 _e(3)) is equal to _______.

Accepted Answer

The given equations are: Hyperbola: 16x^2 - 9y^2 = 144 x^2 9 - y^2 16 = 1 Line: 8x - 3y = 24 y = 8 3 (x - 3) To find the points of intersection, substitute y from the line equation into the hyperbola equation: 16x^2 - 9 ( 8 3 (x - 3) )^2 = 144 16x^2 - 64(x^2 - 6x + 9) = 144 x^2 - 4(x^2 - 6x + 9) = 9 -3x^2 + 24x - 45 = 0 x^2 - 8x + 15 = 0 (x - 3)(x - 5) = 0 x = 3, x = 5 For x [3, 5], the upper curve is the hyperbola y = 4 3 x^2 - 9 and the lower curve is the line y = 8 3 (x - 3). The area A of th

Question 7

Let P_ 1 : y=4 x^ 2 and P_ 2 : y=x^ 2 +27 be two parabolas. If the area of the bounded region enclosed between P_ 1 and P_ 2 is six times the area of the bounded region enclosed between the line y= x,

Accepted Answer

Area bounded between P_1 & P_2 is _ -3 ^ 3 ((x^2 + 27) - (4x^2)) \, dx (P.O.I. of P_1 & P_2 is x = 3) = 2 _ 0 ^ 3 (27 - 3x^2) \, dx = 2[27x - x^3]_ 0 ^ 3 = 2[81 - 27] = 108 Area bounded between P_1 & L is 18 sq. units (Area between x^2 = 4ay & line x = my) is 8a^2 3m^3 Area between x^2 = y 4 & x = y is 8 ( 1 16 )^2 3 ( 1 )^3 = 18 8 16 16 3 ^3 = 18 ^3 = 2^6 3^3 = 12

Question 8

The area of the region R = \ (x, y): x y 8,1 y x^ 2 , x 0 \ is

Accepted Answer

The region R is defined by xy 8, 1 y x^2, and x 0. First, find the intersection points of the curves. Intersection of y = 1 and y = x^2: x^2 = 1 x = 1 (since x 0). Intersection of y = x^2 and xy = 8: x(x^2) = 8 x^3 = 8 x = 2. At x=2, y=4. Intersection of xy = 8 and y = 1: x(1) = 8 x = 8. The region is bounded below by y = 1 from x = 1 to x = 8. The upper boundary consists of two parts: From x = 1 to x = 2, the upper curve is y = x^2. From x = 2 to x = 8, the upper curve is y = 8 x . The area A i

Question 9

Let f( ) denote the area of the region in the first quadrant bounded by x=0, x=1, y^ 2 =x and y=| x-5|-|1- x|+ x^ 2 . Then (f(0)+f(1)) is equal to

Accepted Answer

For = 0: curve is y = 5 - 1 = 4 (horizontal line), area = _0^1 (4 - x ) dx = 4 - 2 3 = 10 3 For = 1 and 0 x 1: |x-5| = 5-x, |1-x| = 1-x, so y = 4 + x^2 Area = _0^1 (4 + x^2 - x ) dx = 4 + 1 3 - 2 3 = 11 3 f(0) + f(1) = 10 3 + 11 3 = 7

Question 10

Let A _ 1 be the bounded area enclosed by the curves y=x^ 2 +2, x+y=8 and y-axis that lies in the first quadrant. Let A _ 2 be the bounded area enclosed by the curves y=x^ 2 +2, y^ 2 =x, x=2, and y-ax

Accepted Answer

For A_1: intersection of y = x^2 + 2 and x + y = 8 at x = 2, y = 6 A_1 = _0^2 (8-x-(x^2+2)) dx = _0^2 (6-x-x^2) dx = [6x - x^2 2 - x^3 3 ]_0^2 = 12 - 2 - 8 3 = 22 3 For A_2: bounded by y = x^2 + 2, y^2 = x (i.e., x = y^2), x = 2, and y-axis. In first quadrant, y^2 = x goes from (0,0) to (2, 2 ). The parabola y = x^2 + 2 passes through (0,2) and (2,6). A_2 = _0^ 2 (2 - y^2) dy = [2y - y^3 3 ]_0^ 2 = 2 2 - 2 2 3 = 4 2 3 A_1 - A_2 = 22 3 - 4 2 3 = 2 3 (2 2 +1)

Question 11

The area of the region enclosed between the circles x^ 2 +y^ 2 =4 and x^ 2 +(y-2)^ 2 =4 is:

Accepted Answer

Circle 1: centre (0,0), r=2. Circle 2: centre (0,2), r=2. Distance between centres d=2. Intersection: y^2 = (y-2)^2 y = 1, giving x = 3 . Angle subtended at centre of circle 1 by the chord: = 2 /3. Segment area = 1 2 (4) ( 2 3 ) - 1 2 (4) 2 3 = 4 3 - 3 . By symmetry, total intersection area = 2 ( 4 3 - 3 ) = 8 - 6 3 3 = 2 3 (4 - 3 3 ).

Question 12

Let the area of the region bounded by the curve y= \ x, x\ , lines x=0, x= 3 2 , and the x-axis be A. Then, A + A ^ 2 is equal to \_\_\_\_.

Accepted Answer

A = _ 0 ^ /4 x \, dx + _ /4 ^ x \, dx + _ ^ 5 /4 - x \, dx + _ 5 /4 ^ 3 /2 - x \, dx A = ( x)_ 0 ^ /4 + (- x)_ /4 ^ + ( x)_ ^ 5 /4 + ( x)_ 5 /4 ^ 3 /2 A = 1 2 + 1 2 + 1 + 1 - 1 2 + 1 - 1 2 = 3 A^2 + A = 12

Question 13

The area of the region A = \ (x, y): 4 x^ 2 +y^ 2 8 . and .y^ 2 4 x \ is:

Accepted Answer

A = _ 0 ^ 2 2 x \, dx + 2 _ 1 ^ 2 8 - 4x^2 \, dx = 8 3 ( x^ 3 2 ) |_ 0 ^ 1 + 4 _ 1 ^ 2 2 - x^2 \, dx = 8 3 + 4 1 2 [ x 2 - x^2 + 2 ^ -1 ( x 2 ) ] |_ 1 ^ 2 = 8 3 + 2 [ 2 2 - 1 - 2 4 ] = 8 3 + 2 - 2 - = + 2 3 sq. units

Question 14

Let the line x=-1 divide the area of the region \ (x, y): 1+x^ 2 y 3-x \ in the ratio m: n, gcd (m, n)=1. Then m+n is equal to

Accepted Answer

m n = _ -1 ^ 1 [(3-x) - (1+x^2)] \, dx _ -2 ^ 1 [(3-x) - (1+x^2)] \, dx = 20 7 m + n = 20 + 7 = 27

Question 15

If the area of the region \ (x, y): 1-2 x y 4-x^ 2 , x 0, y 0 \ is , , N , gcd ( , )=1, then the value of ( + ) is :

Accepted Answer

Required area = 2 3 8 - 1 2 1 2 1 = 16 3 - 1 4 = 61 12 = + = 73

Question 16

The area of the region, inside the ellipse x^ 2 +4 y^ 2 =4 and outside the region bounded by the curves y=|x|-1 and y=1-|x|, is :

Accepted Answer

Ellipse: x^2 4 + y^2 = 1 with a = 2, b = 1. Area = ab = 2 . Curves y = |x| - 1 and y = 1 - |x| intersect at x = 1, forming a rhombus with vertices (0, 1), (1, 0), (0, -1), (-1, 0). Vertices (0, 1) lie on ellipse; ( 1, 0) are inside ellipse. Area of rhombus = 1 2 2 2 = 2. Required area = 2 - 2 = 2( - 1).

Area Under Curves — Previous Year Questions & Solutions