JEE Advanced (Single Correct Type): Definite Integrals & Application of Integrals | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1.
(a)
(b)
(c)
(d) None of these

Correct Answer is option (b)

Since

Q.2. The area bounded by the curves y² = 4x and y = x is equal to
(a) 1/3
(b) 8/3
(c) 35/6
(d) None of these

Correct Answer is option (b)

For the given curves, y² = 4x and y = x, the intersection points are (0, 0) and (4, 4).

Therefore, the area bounded by the curves,

A =0∫4 [√(4x)-x] dx [since y² = 4x, y = √(4x) ]

A = 2. 0∫4√x dx – 0∫4 x dx

Integrate the function and apply the limits, we get

A = (4/3)(8) – 8

A = (32-24)/3 = 8/3.

Hence, option (b) is the correct answer.

Q.3.
(a) 50
(b) 24
(c) 20
(d) None of these

Correct Answer is option (c)

Q.4. The area of the figure bounded by the curve y = log_ex, the x-axis and the straight line x = e is
(a) 5 - e
(b) 3 + e
(c) 1
(d) None of these

Correct Answer is option (c)
At, x= 1, y = log_e (1) = 0
At, x = e, y = log_e (e) = 1
Therefore, A = ₁∫^e log_ex dx
Using integration by parts,
A = [x log_e x – x]₁^e
Now, apply the limits, we get
A = [e-e-0+1]
A = 1

Q.5.
(a) π|
(b) π/2|
(c) π/4|
(d) π/3| 

Correct Answer is option (c)

Now adding (i) and (ii), we get

Q.6. The area of the region bounded by the curve x² = 4y and the straight line x = 4y – 2 is
(a) ⅜ sq. units
(b) ⅝ sq. units
(c) ⅞ sq. units
(d) 9/8 sq. units

Correct Answer is option (d)

For the curves x² = y and x = 4y-2, the points of intersection are x = -1 and x = 2.

Hence, the required area, A = _-1∫² {[(x+2)/4]- [x²/4] } dx

Now, integrate the function and apply the limits, we get

A = (¼)[(10/3)-(-7/6)]

A = (¼)(9/2) = 9/8 sq. units

Hence, the correct answer is option (d) 9/8 sq. units

Q.7.
(a) π|
(b) π/2|
(c) π/3|
(d) π/4| 

Correct Answer is option (d)

 On adding,  

Q.8. The area enclosed between the graph of y = x³ and the lines x = 0, y = 1, y = 8 is
(a) 7
(b) 14
(c) 45/6
(d) None of these

Correct Answer is option (c)

Given curve, y=x³ or x = y^1/3.

Hence, the required area, A = ₁∫⁸ y^1/3 dy

A = [(y^4/3)/(4/3)]18

Now, apply the limits, we get

A = (¾)(16-1)

A = (¾)(15) = 45/4.

Hence, option (c) 45/4 is the correct answer.

Q.9.  if
(a) e^x(x³+3x²) 
(b) x³e^x
(c) a³e^a
(d) None of these

Correct Answer is option (b)

Q.10.  The area of the region bounded by the curve y² = x, the y-axis and between y = 2 and y = 4 is
(a) 52/3 sq. units
(b) 54/3 sq. units
(c) 56/3 sq. units
(d) None of these

Correct Answer is option (c)

Given: y² = x

Hence, the required area, A = ₂∫⁴ y² dy

A = [y³/3]2⁴

A = (4³/3) – (2³/3)

A = (64/3) – (8/3)

A = 56/3 sq. units.

Q.11.  
(a) 1
(b) 0
(c) 2

(d) -2

Correct Answer is option (b)
 Let f(x)=x|x|. Then f(−x)=−x|−x|=−x|x|=−f(x)
Therefore
   
(By the property of definite integral).

Q.12.  Area of the region bounded by the curve y = cos x between x = 0 and x = π is
(a) 1 sq. units
(b) 2 sq. units
(c)  3 sq. units
(d) 4 sq. units

Correct Answer is option (b)

Given: y= cos x and also provided that x= 0 and x = π

Hence, the required area, A =₀∫^π |cos x| dx

It can also be written as,

A = 2₀∫^π/2 cos x dx

Now, integrate the function, we get

A = 2[sin x]₀^π/2

Now, apply the limits we get

A = 2 sq. units

Hence, the correct answer is option (b) 2 sq. units

Q.13.
(a)
(b)
(c)

(d)

Correct Answer is option (b)
     ...(i)
      ....(ii)
  By adding (i) and (ii), we get

Q.14.  Area of the region bounded by the curve x = 2y + 3, the y-axis and between y = -1 and y = 1 is
(a) 6 sq. units
(b) 4 sq. units
(c) 8 sq. units
(d) 3/2 sq. units

Correct Answer is option (a)
Required Area =_-1∫¹(2y+3)dy

A=[(2y²/2)+3y]_-1¹

Now, apply the limits, we get

A = 1+3-1+3

A = 6 sq. units.

Q.15.   
(a)
(b)
(c)  πlog_e2 

(d) 0

Correct Answer is option (d)

Q.16. The area bounded by the curve y = x3, the x-axis and two ordinates x = 1 and x = 2 is
(a) 15/2 sq. units
(b) 15/4 sq. units
(c) 17/2 sq. units
(d) 17/4 sq. units

Correct Answer is option (b)

Required Area = ₁∫² x³ dx

A = [x⁴/4]₁²
Now, apply the limits, we get
A = [(2⁴/4) – (¼)]
A = (16/4) – (¼)
A = 15/4
Hence, the required area is 15/4 is the correct answer.

Q.17. 
(a)
(b)  
(c)  
(d)  

Correct Answer is option (a)




 (Putting 2x=t)  
 

Q.18. The area of the region bounded by the circle x² + y² = 1 is
(a) 2π sq. units
(b) 3π sq. units
(c) 4π sq. units
(d) 1π sq. units

Correct Answer is option (d)
Given:
Given circle equation is x²+y² =1, whose centre is (0, 0) and radius is 1.
Therefore, y² = 1-x²
y=√(1-x²)
Hence, the required area, A = 4₀∫¹√(1-x²) dx

Now, integrate the function and apply limits, we get
A = 4(½)(π/2)
A = π sq. units
Hence, option (d) 1π sq. unit is the correct answer.

Q.19.

(a) 2
(b) -2
(c) 0
(d) None of these

Correct Answer is option (c)

Now,

On adding, 2I = 0 ⇒ I = 0.

Q.20. Area bounded by the curve y = sin x and the x-axis between x = 0 and x = 2π is
(a) 2 sq. units
(b) 3 sq. units
(c) 4 sq. units
(d) None of these

Correct Answer is option (c)

Required Area, A = ₀∫²π |sin x| dx
A = ₀∫^π sin x dx + _π∫²π(-sin x) dx

Now, substitute the limits, we get

A= 4 sq. units.