Q: The area of the region R = \ (x, y): 5 x^2 y 2 x^2+9 \ is :

aligned & Required area =2 _0^ 3 (2 x^2+9-5 x^2 ) d x \\ & =2 [9 x-x^3 ]_0^ 3 =2[9 3 -3 3 ]=12 3 . aligned

Question 1

The line y=m x bisects the area unclosed by lines x=0, y=0 and x= 3 2 and the curve y=1+4 x-x^2. Then, the value of m is

Accepted Answer

Given, y=1+4 x-x^2 aligned & d y d x =4-2 x=0 x=2 [point of maxima] \ & y_ =1+4 2-4=5 aligned when, x=0, y=1 Graph according to question and above information aligned & If y=m x bisects the area bounded \ & A_1=A_2 \ & _0^ 3 / 2 (1+4 x-x^2 ) d x=2 _0^ 3 / 2 m x d x \ & [x+2 x^2- x^3 3 ]_0^ 3 / 2 = [m x^2 ]_0^ 3 / 2 \ & 3 2 +2 9 4 - 27 8 1 3 =m 9 4 \ & m= 13 6 aligned

Question 2

If a, c, b are in GP, then the area of the triangle formed by the lines a x+b y+c=0 with the coordinates axes is equal to

Accepted Answer

Given a, c, b are in GP c^2=a b...(i) According to question, aligned Area of O A B & = | 1 2 O A O B | \ = & | 1 2 ( -c b ) ( -c a ) | \ & = 1 2 c^2 a b = 1 2 [by Eq. (i)] aligned

Question 3

The area enclosed by the curves y=x^3 and y= x is

Accepted Answer

Given curves are y=x^3 and y= x Both curves intersects at x=0 and x=1 Shaded region is required area aligned A & = | _0^1 (x^3- x ) d x |= | [ x^4 4 - 2 x^ 3 / 2 3 ]_0^1 | \ & = | [ 1 4 - 2 3 ] | \ & = 5 12 sq. unit aligned

Question 4

The area of the region bounded by the curves x =y^2-2 and x=y is

Accepted Answer

Given, x=y^2-2 and x=y. On solving, x=y^2-2 and x=y, we get (-1,-1) and (2,2). Area of the shaded region, aligned & A= _ -1 ^2 y d y- _ -1 ^2 (y^2-2 ) d y \ & = [ y^2 2 - y^3 3 +2 y ]_ -1 ^2= ( 4 2 - 8 3 +4 )- ( 1 2 + 1 3 -2 ) \ & = 10 3 + 7 6 = 27 6 = 9 2 . aligned

Question 5

If the area bounded by the curves y=a x^2 and x =a y^2,(a 0) is 3 sq units, then the value of a is

Accepted Answer

We have given, y=a x^2 ....(i) and x=a y^2 ......(ii) Put the value of y by Eq. (i) in Eq. (ii), we get x=a a^2 x^4 x^4 a^3-x=0 x (x^3 a^3-1 )=0 x=0, 1 2 When, x=0 y=0 and x= 1 a y= 1 a Here, points of intersection of curves y=a x^2 and x=a y^2 are (0,0) and ( 1 a , 1 a ). Required area A= _ x=a ^ x=b [f_2(x)-f_1(x) ] d x 3= _0^ 1 / a ( x a -a x^2 ) d x 3= [ 1 a 2 3 x^ 3 / 2 - a x^3 3 ]_0^ 1 / a 3= 2 3 a [ ( 1 a )^ 3 / 2 ]- a 3 [ ( 1 a )^3 ] 3= 2 3 a 1 a a - a 3 1 a^3 3= 2 3 a^2 - 1 3 a^2 3= 2-1

Question 6

The area of the region R = \ (x, y): 5 x^2 y 2 x^2+9 \ is :

Accepted Answer

aligned & Required area =2 _0^ 3 (2 x^2+9-5 x^2 ) d x \ & =2 [9 x-x^3 ]_0^ 3 =2[9 3 -3 3 ]=12 3 . aligned

Question 7

The area enclosed between the curve y= _e(x+e) and the coordinate axes is

Accepted Answer

Required area A= _ 1-e ^0 y d x= _ 1-e ^0 _e(x+e) d x put x+e=t d x=d t also when x=1-e, t= 1 and when x=0, t=e A= _1^e _e t d t= [t _e t-t ]_1^e e-e-0+1=1

Question 8

The area of the region bounded by the parabola (y-2)^2=(x-1), the tangent to the parabola at the point (2,3) and the X-axis is

Accepted Answer

The equation of the parabola is y^2-4 y-x+5=0 The equation of tangent at (2,3) is aligned 3 y-2(y+3)- x+2 2 +5 & =0 \ 2 y-x-4 & =0 aligned Required area A is given by aligned & A= _0^3 (x_2-x_1 ) d y \ & A= _0^3 [ \ (y-2)^2+1 \ -\ 2 y-4\ ] d y \ & A= _0^3 (y^2-6 y+9 ) d y= _0^3(3-y)^2 d y \ & =- [ (3-y)^3 3 ]_0^3=9 \ & aligned

Question 9

The area enclosed by the curves y= x+ x and y=| x- x| over the interval [0, 2 ] is

Accepted Answer

Given, y= x+ x, x [0, / 2] d y d x = x- x aligned y & =| x- x| \ & = [ array cc x- x & x [0, / 4] \ x- x & x [ / 4, / 2] array ] aligned Required area aligned &= _0^ / 4 |( x+ x)-( x- x)| d x+ _ / 4 ^ / 2 |2 x| d x aligned = _0^ / 4 |2 x| d x+ _ / 4 ^ / 2 |2 x| d x =2(- x)_0^ / 4 +2( x)_ / 4 ^ / 2 =2 [- 1 2 +1+1- 1 2 ] aligned = & 2 (2- 2 2 ) \ = & 2(2- 2 ) \ = & 4-2 2 \ = & 2 2 ( 2 -1) aligned

Question 10

The area bounded by the curves x+2 y^ 2 =0 and x+3 y^ 2 =1 is

Accepted Answer

We have, x+2 y^ 2 =0 y^ 2 =- x 2 (i) a parabola with vertex (0,0) and x+3 y^ 2 =1 y^ 2 = 1-x 3 =- ( x-1 3 ) (ii), a parabola with vertex (1,0) Solving (i) and (ii), we get y= 1 A = _ -1 ^ 1 ( (1-3 y^ 2 )- (-2 y^ 2 ) ) d y =2 _ 0 ^ 1 (1-y^ 2 ) d y=2 [y- y^ 3 3 ]_ 0 ^ 1 = 4 3 sq.units

Question 11

The area bounded by curves (x-1)^ 2 +y^ 2 =1 and x^ 2 +y^ 2 =1 is

Accepted Answer

Given circles are x^ 2 +y^ 2 =1 (i) and (x-1)^ 2 +y^ 2 =1 (ii) Centre of (i) is O (0,0) and radius =1 Both these circle are symmetrical about x-axis solving (i) and (ii), we get, -2 x+1=0 x= 1 2 array l then y^ 2 =1- ( 1 2 )^ 2 =34 \ y= 3 2 array The points of intersection are P ( 1 2 , 3 2 ) and Q ( 1 2 ,- 3 2 ) It is clear from the figure that the shaded portion in region whose area is required. Required area = area OQAPO =2 area of the region OLAP =2 ( area of the region O L P O+ area of L A

Question 12

The line y=m x bisects the area enclosed by lines x =0, y =0 and x =3 / 2 and the curve y =1+4 x - x ^ 2 Then the value of m is

Accepted Answer

y=1+4 x-x^ 2 =5-(x-2)^ 2 We have _ 0 ^ 3 / 2 (1+4 x-x^ 2 ) dx =2 _ 0 ^ 3 / 2 mx dx = 3 2 +2 ( 9 4 )- 1 3 ( 27 8 )= m 9 4 On solving we get m = 13 6

Area Under Curves — Previous Year Questions & Solutions