# Application of Integrals - Online Test

Q1. The area of ellipse  is
Explaination / Solution:

Area of standard ellipse is given by :πab.

Q2. The area of the loop between the curve y = a sin x and the x – axis and x= 0 , x=π. is
Explaination / Solution:

Required area =
0πasinxdx=a[cosx]π0=a(cosπ+cos0)=a(1+1)=2a

Q3. The area enclosed between the curves y = ,x- axis and two ordinates x = 1 to x = 2 is [ in square units ]
Explaination / Solution:

Required area :
= == sq units

Q4. Area of the region bounded by the curves y = , x = a , x = b and the x- axis is given by
Explaination / Solution:

Required area
abexdx=[ex]ba=ebea

Q5. 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
Explaination / Solution:

Required area :
area of ellipse – area of right angled triangle AOB.
= ab –  ab =  ( - 2 ) .

Q6. The area lying in the first quadrant and bounded by the curve y =  , the x – axis and the ordinates at x = - 2 and x = 1 is
Explaination / Solution:

Required area :

Q7. The area bounded by the curve y =  , the x – axis and two ordinates x = 1 and x = 2 is
Explaination / Solution:

sq.units.

Q8.

The area bounded by y = 2cosx , x = 0 to x =2π and the axis of x in square units is

Explaination / Solution:

$= == 2 Therefore , total area from x= 0 to x = 2 is 4 X 2= 8 sq. units. Q9. The area of the figure bounded by the curve , the x – axis and the straight line x = e is Answer : Option D Explaination / Solution: Required area :$ = - = 1

Q10.

The area enclosed by the curve  is

Explaination / Solution:

Required area :
\$= ==