Application of Integrals - Online Test
Q1. The area of ellipse is
Answer : Option D
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
Answer : Option D
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 ]
Answer : Option C
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
Answer : Option A
Explaination / Solution:
Required area
∫abexdx=[ex]ba=eb−ea
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
Answer : Option B
Explaination / Solution:
Required area :
= area of ellipse – area of right angled triangle AOB.
= 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
Answer : Option D
Explaination / Solution:
Required area :
Q7. The area bounded by the curve y = , the x – axis and two ordinates x = 1 and x = 2 is
Answer : Option B
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
Answer : Option C
Explaination / Solution:
$= = == 2
Therefore , total area from x= 0 to x = 2 is 4 X 2= 8 sq. units.
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
$= = - = 1
Q10.
The area enclosed by the curve is
Answer : Option A
Explaination / Solution:
Required area :
$= = == =
$= = == =
