Solved Examples on Area Bounded by the curve - Study Material for IIT JEE

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Solved Examples of Definite Integral Part I

11:     The area bounded by the curve y = x (3 – x)2 , the x – axis and the ordinates of the maximum and minimum points of the curve is

(A)       2 sq. units                      

(B)       4 sq. units

(C)       3 sq. units                      

(D)       8 sq. units


	
		
			
			Solution:             y = x (3 – x)2

			                                    After solving , we get x =1 and x = 3         are       points of maximum and     minimum respectively.

			                              Now the shaded region is the required region

			                              
			
			
			 
			
		
	


12:                        The area enclosed by the curve |x + 1| + |y + 1| = 2 is

(A)       3 sq. units                       (B)       4 sq. units

(C)       5 sq. units                       (D)       8 sq. units


	
		
			
			Solution:             Shift the origin at the point (-1, -1)

			                                    So |x + 1| + |y + 1| = 2 

			                                    becomes |x| + | y | = 2

			                                    Hence required area is

			                                    = 4 × 1/2 × 2 × 2 = 8sq. units
			
			
		
	


13:                             The area common to the curves y = x3 and y =  √x is

(A)       2                                        (B)       4

(C)       8                                        (D)       None of these


	
		
			
			 
			
			
			 

			
			
		
	


14:       The area of the region consisting of points (x, y) satisfying |x ± y | < 2 and x2 + y2 > 2 is

            (A)             8 – 2π sq. units

            (B)             4 – 2π sq. units

            (C)             1 – 2π sq. units

            (D)             2π sq. units


                                  


Solution:            Shaded region is the required one.

                          Required Area   =  4 × 1/2 × 2 × 2 – π.2 = 8 – 2π sq. unit

15:                   ∫xo [sint] dt where x ∈ and [2nπ, (4n+1)]π, n ∈ N and [.] denotes the greatest integer function is equal to.

                        (A)       -nπ                                         (B)       -(n +1)π

                        (C)       -2nπ                                      (D)       -2(n +1)π



16:                  If f(π) = 2 and ∫πo [f(x) + f"(x)] sin x dx = 5 then f(0) is equal to, (it given that f(x) is continuous in [0, π])

                        (A)       7                      (B)       3                      (C)       5                      (D)       1



                    => f(π) + f(0) = 5 (given)

                    => f(0) = 5 – f(π) = 5 – 2 = 3

17:                Let f(x) is a continuous function for all real values of x and satisfies  + a then value of ‘a’ is equal to

                        (A)  –1/24                   (B)      17/168                (C)      1/7        (D)      None of these



                        Diff. both sides of (i) w. r. t. x we get;

                        f(x) = 0 – x2 f(x) + 2x15 + 2x5



18:                  . Then complete set of values of x for which f(x) <lnx is

                        (A)       (0, 1]                                       (B)       [1, ∞)

                        (C)       (0, ∞)                                      (D)       None of these

Solution:       

                        Let g(x) = f(x) - ln(x). x ∈ R +

                        => g ’(x) = f ’(x) – 1/x = ex – 1 / x > 0 ∀ x ∈ R+

                        => g'(x) is increasing for ∀ x ∈ R+

                        g(1) = f(1) - ln1 = 0 – 0 = 0

                        => g(x) > 0 ∀ x > 1 and g(x) > 0      ∀ x ∈ (0, 1]

                        => ln x > f(x) ∀ x ∈ (0, 1]

19:                  , where [.] denotes the greatest integer function, and x ∈ R+, is equal to

                        (A)       1/In2 ([x] + 2[x]–1)            (B)       1/In2 ([x] + 2[x])

                        (C)       1/In2 ([x] – 2[x])               (D)       1/In2 ([x] + 2[x]+1)

Solution:       Let n < x < n + 1 where n ∈ I, I > 0


              


20:                  Let I1 =  then

                        (A)      I1 > I2 > I3 > I4                         (B)      I2 > I3 > I4 > I1

                        (C)      I3 > I4 > I1 > I2                         (D)      I2 > I1 > I3 > I4



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