Application of Integrals - Online Test

Answer : Option B
Explaination / Solution:

 y = 4−x2>0⇒−2<x<2  And  y=4−x2<0⇒x<−2or...x>2
Required area:
 = ∫02(4−x2)dx+∣∣∣∫23(4−x2)dx∣∣∣= 163+∣∣−73∣∣ =233sq.units

Answer : Option B
Explaination / Solution:

The given curve consists of two straight lines x + y = 1 ( x ≥ 0 )and -x + y = 1 ( x < 0 )
Required area :
= 2∫01ydx = 2∫01(1−|x|)dx=2∫01(1−x)dx=2[x−x22]10 = 1sq.unit 

Answer : Option A
Explaination / Solution:

The area of the standard ellipse is given by ;πab. Here , a = 5 and b = 4 Therefore , the area of curve is π(5)(4)=20π.

Answer : Option C
Explaination / Solution:

Given equation is of a circle ; (x−a)2+y2=a2 Its radius is equal to a. Therefore required area is = πa2.

Answer : Option C
Explaination / Solution:

For x – axis , y = 0,
Therefore, 4x − x2= 0
Therefore , x = 0 or x = 4.
Required area :
 ∫04y.dx = ∫04(4x−x2).dx= [2x2−x33]40 = 32 - 643- 0 = 323

Answer : Option B
Explaination / Solution:

Required area :

= ∫04(y2−4)dy= [y33−4y]40=163sq.units

Answer : Option C
Explaination / Solution:

Required area :
 = ∫12y.dx =∫12logxdx =[xlogx−x]21= 2log2−2−log1+1= 2log2−1 = 2log2−loge= log4−loge =  log(4/e)

Answer : Option A
Explaination / Solution:

Answer : Option B
Explaination / Solution:

The two curves parabola and the line meet where,
 3 - x =x2+1 ⇒x2+x−2 = 0 ⇒x=−2,1.$\mathrm{.}$
Required area ;
∫−21{3−x−(x2+1)}dx = [2x−x22−x33]1−2= 92

Answer : Option A
Explaination / Solution:

The two curves meet in ( 0 , 0 ) and ( 1, 1 ).The required area lies above the curve y = x2 and below x = y2 and is equal to ;