21:       Area bounded by y = g(x), x-axis and the lines x = -2, x = 3, where

                          

                       and f(x) = x² -  , is equal to

                        (A)  113/24 sq.units                       (B)         111/24 sq.units

                        (C)  117/24 sq.units                       (D)         121/24 sq,units

22:                  Area of the region which consists of all the points satisfying the conditions |x–y| + |x+y| < 8 and xy > 2, is equal to

                        (A)      4(7 – ln8)sq. units               (B)      4(9 – ln8)sq. units  

                        (C)      2(7 – ln8)sq. units               (D)      2(9 – ln8)sq. units

Solution:       The expression |x–y| + |x+y| < 8, represents the interior region of the square formed by the lines x = ± 4, y = ± 4 and xy > 2. represents the region lying inside the hyperbola xy =2

                        Required area

                      

23:                  Area bounded by the parabola (y - 2)² = x – 1, the tangent to it at the point P (2, 3) and the x-axis is equal to

                        (A)      9 sq. units                             (B)      6 sq. units

                        (C)      3 sq. units                             (D)      None of these

Solution:       (y - 2)² = (x - 1) => 2(y - 2).  = 1

                                 

                        => dy/dx = 1/2(y–2)

                        Thus equation of tangent at P(2, 3) is,

                        (y – 3) = 1/2 (x–2) i.e., x = 2y – 4

                        Required area 

                        = ((y–2)³/3 – y² + 5y)³₀ = 9 sq. units

24:                  Two lines draw through the point P(4, 0) divide the area bounded by the curves y = √2 πx/4 and x – axis, between the linea x = 2, and x = 4, in to three equal parts. Sum of the slopes of the drawn lines is equal to

                        (A) –2 2/π                                     (B)      –√2/π

                        (C) –√2/π                                     (D)      –4√2/π

Solution:       Area bounded by y = √2 .sin πx/4 and x-axis between the lines x = 2 and x = 4,

                                    

                      Let the drawn lines are L₁: y – m₁(x - 4) = 0 and L₂: y – m₂(x - 4) = 0, meeting the line x = 2 at the points A and B respectively Clearly A = (2, - 2m₁); B= (2, -2m₂)

                        (A)   1/π+2 – A                               (B)      1/2 + 1/π+2 – A

                        (C)   1/2 – 1/π+2 – A                     (D)      1/2 + 1/π+2 + A 

26.                   The value of the integral  is

                        (A) 1                                                                         (B) π/12

                        (C) π/6                                                                     (D) none of these  

Solution:       Using the property  f (a + b – x) dx, the given integral 

                        Hence (B) is the correct answer.

27.                   If I =   dx then

                        (A) 0                                                               (B) 2  

                        (C)  π/2                                                         (D) 2 – π/2

                        Hence (D) is the correct answer.

28.                   If I =  , then

                        (A) 0 < I < 1                                                   (B) I > π/2

                        (C) I < √2π                                                     (D) I > 2 π

Solution:       Since x ∈ [0, π/2]  => 1 < 1 + sin³ x < 2

                        Hence (C) is the correct answer.

29.                   If f (a + b –x) = f (x) then ∫^b_a x f (x) dx is equal to

                        (A) a–b/2 ∫^b_a f(x) dx                                    (B) (a+b/2) ∫^b_a f(x) d x           

                        (C) 0                                                              (D) none of these

Solution:        I = ∫^b_a x f (x) dx = ∫^b_a (a+b-x) f (a +b-x) dx

                        = (a + b) ∫^b_a  f(a +b –x) - ∫^b_a x f (a + b –x) dx 

                        = ( a + b) ∫^b_a f (x) dx - ∫^b_a x f (x) d x

                        Hence I = (a+b/2) ∫^b_a f(x) dx.

                        Hence (B) is the correct answer.

To read more, Buy study materials of Definite integral comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.