Applications of the Integrals Test

Result

Applications of the Integrals Test - 3

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

The area enclosed between the curves y = x³ ,x- axis and two ordinates x = 1 to x = 2 is [in square units]

A
15/4

B
5/3

C
5/4

D
12/5

Solution
Question 2
1 / -0.25

The area bounded by y = log x , the x –axis and the ordinates x = 1 and x = 2 is

A
log 2/e

B
log 4

C
log 4/e

D
none of these

Solution
Question 3
1 / -0.25

The area bounded by the curves y² = x³ and |y| = 2x is given by

A
2

B
32/5

C
32/2

D
none of these

Solution

Both y² = x³ and |y| = 2x are symmetric about y –axis and on solving them we get :
Question 4
1 / -0.25

The area bounded by the curve y = 2x - x² and the line x + y = 0 is

A
9/2 sq. units

B
35/6 sq. units

C
19/6 sq. units

D
none of these

Solution

The equation y = 2x − x² i.e. y –1 = - (x - 1)² represents a downward parabola with vertex at (1, 1) which meets x –axis where y = 0 .i .e . where x = 0 , 2. Also , the line y = - x meets this parabola where –x = 2x − x² i.e. where x = 0 , 3.
Therefore , required area is :
Question 5
1 / -0.25

The area bounded by the parabola y = x² and the line y = x is

A
1/2 sq. units

B
1/3 sq. units

C
1/6 sq. units

D
none of these

Solution
Question 6
1 / -0.25

Area of the region bounded by the curves y = e^x , x = a , x = b and the x- axis is given by

A
e^b −e^a

B
e^{b −a}

C
e^b +e^a

D
none of these

Solution
Question 7
1 / -0.25

The area bounded by the curves y² = 4x and y = x is equal to

A
8/3

B
35/6

C
1/3

D
none of these

Solution

The two curves y² = 4x and y = x meet where x² = 4x i.e ..where x = 0 or x = 4 . Moreover , the parabola lies above the line y = x between x = 0 and x = 4 . Hence , the required are is :
Question 8
1 / -0.25

The area bounded by the curves y = √x, 2y+3 = x and the x- axis in the first quadrant is

A
9

B
36

C
25

D
none of these

Solution

To find area the curves y = √x and x = 2y + 3 and x –axis in the first quadrant., We have ; y² −− 2y −− 3 = 0 (y –3) (y + 1) = 0 . y = 3 , - 1 . In first quadrant , y = 3 and x = 9.
Therefore , required area is ;
Question 9
1 / -0.25

Let y be the function which passes through (1 , 2) having slope (2x + 1) . The area bounded between the curve and the x –axis is

A
1/6 sq. units

B
6 sq. units

C
5/6 sq. units

D
none of these

Solution

Given slope of the curve is 2x + 1.

Also , it passes through (1, 2).
∴2 = 1+1+c ⇒c = 0
Equation of curve is : y = x² + x. Therefore , points of intersection of y = x (x +1) and the x –axis are x = 0 , x = - 1.
Required area :
Question 10
1 / -0.25

AOB is the positive quadrant of the ellipse in which OA = a and OB = b . The area between the arc AB and chord AB of the ellipse is

A
(1/2) ab(π+2)

B
(1/4) ab(π-2)

C
(1/4) ab(π-4)

D
none of these

Solution

Required area :
= (1/4) area of ellipse –area of right angled triangle AOB. (π - 2)
Question 11
1 / -0.25

The area common to the circle x² +y² = 16 and the parabola y² = 6x is

A

B

C

D
none of these

Solution

Solving eqns. (i) and (ii), we get points of intersection (2, 2 √3) and (2, -2 √3)

Substituting these values of x in eq. (ii). Since both curves are symmetrical about r-axis.

Hence the required area
Question 12
1 / -0.25

The area included between the curves is equal to

A
1/6

B
4/3

C
2/3

D
none of these

Solution

Required area :
Question 13
1 / -0.25

The area bounded by the curves y² = 20x and x² = 16y is equal to

A
80 π sq. units

B
100 π sq. units

C
320/3 sq. units

D
none of these

Solution

Eliminating y, we get :

Required area:
Question 14
1 / -0.25

The area bounded by the curves

A

B

C

D

Solution

Required area :
Question 15
1 / -0.25

The area of the region bounded by the curves y = |x −2|, x = 1 , x = 3 and the x –axis is

A
2

B
1

C
3

D
4

Solution