11. The value of the definite integral ^e∫₁ √x In (x)dx is

1. 1. 4	√e³ +	2
      9		9

   
   
   

   2. 2	√e³ -	4
      9		9

   
   
   

   3. 2	√e³ +	4
      9		9

   
   
   

   4. 4	√e³ -	2
      9		9

   
   
   

---

1. View Hint View Answer Discuss in Forum

   ^e∫₁ √x In xdx = [∠x ∫ √x dx - ∫ ((d/dx) < nx)(√xdx)dx]^es
   (Integrate by parts)

   Correct Option: C

   ^e∫₁ √x In xdx = [∠x ∫ √x dx - ∫ ((d/dx) < nx)(√xdx)dx]^es
   (Integrate by parts)

---

12. The integral ∮_c(ydx - xdy) is evaluated along the circle x² + y² = 1/4 traversed in counter clockwise direction. The integral is equal to

1. 1. 0

   
   
   

   2. - π/4

   
   
   

   3. - π/2

   
   
   

   4. π/4

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given integral ∮_c (ydx - xdy)
   where C is x² + y² = 1/4
   Applying Green’s theorem
   

   ∮cMdx - Ndy = ∫∫R		δN	-	δm		dx dy
   		δx		δx

   
   where R is region included in c
   ∮_cydx - xdy = ∫∫_R(-1-1)dx dy
   = - 2∫∫_R = –2 × Region_R
   = – 2 × area of circle with radius 1/2
   

   = 2 × π		1		²	=	-π
   		2				2

   Correct Option: C

   Given integral ∮_c (ydx - xdy)
   where C is x² + y² = 1/4
   Applying Green’s theorem
   

   ∮cMdx - Ndy = ∫∫R		δN	-	δm		dx dy
   		δx		δx

   
   where R is region included in c
   ∮_cydx - xdy = ∫∫_R(-1-1)dx dy
   = - 2∫∫_R = –2 × Region_R
   = – 2 × area of circle with radius 1/2
   

   = 2 × π		1		²	=	-π
   		2				2

---

---

13. The value of the integral ²∫₀^x∫₀dydx

1. 1. 1	(e - 1)
      2

   
   
   

   2. 1	(e² - 1)²
      2

   
   
   

   3. 1	(e² - e)
      2

   
   
   

   4. 1		e -	1		²
      2			e

   
   
   

---

1. View Hint View Answer Discuss in Forum

   ²∫₀^x∫₀e^e+ydy dx = ²∫₀e^x(^x∫₀e^ydy)dx
   = ²∫₀e^x(e^y)₀^x dx = ²∫₀(e^x - 1)dx
   

   = 2∫0(e2x - ex) dx =		e2x	- ex		²
   		2			0

   
   

   =	e4	- e²	1	+ 1 =	e4	- e²	1
   	2		2		2		2

   
   

   =	1	(e4 - 2e² + 1) =	1	(e² - 1)²
   	2		2

   Correct Option: B

   ²∫₀^x∫₀e^e+ydy dx = ²∫₀e^x(^x∫₀e^ydy)dx
   = ²∫₀e^x(e^y)₀^x dx = ²∫₀(e^x - 1)dx
   

   = 2∫0(e2x - ex) dx =		e2x	- ex		²
   		2			0

   
   

   =	e4	- e²	1	+ 1 =	e4	- e²	1
   	2		2		2		2

   
   

   =	1	(e4 - 2e² + 1) =	1	(e² - 1)²
   	2		2

---

14. The value of the integral
    

    ∞∫-∞	sin x	dx is
    	x² + 2x + 2

    
    evaluated using contour integration and the residue theorem is

1. 1. πsin (1)

      e

   
   
   

   2. - πcos (1)

      e

   
   
   

   3. sin (1)

      e

   
   
   

   4. cos (1)

      e

   
   
   

---

1. View Hint View Answer Discuss in Forum

   I = ∞∫-∞	sin (x)
   	x² + 2x + 2

   
   

   let f(x) =	Im(eix)
   	z² + 2z + 2

   
   Then poles of f(z) are given by z² + 2z + 2 = 0
   ∴ z = – 1 ± i
   R₁ = Res: (f(z): z = -1 + i)
   

   Ltz→-1±1[z-(-1+i)]	eix
   	[z-(-1+1)][z-(-1-i)]

   
   

   =	ei(-1+i)	=	e-i-1
   	-1 + i + 1 + i		2i

   
   
   

   = IM[πe-1(cos(1) - isin(1))] = -	π sin(1)
   	e

   Correct Option: A

   I = ∞∫-∞	sin (x)
   	x² + 2x + 2

   
   

   let f(x) =	Im(eix)
   	z² + 2z + 2

   
   Then poles of f(z) are given by z² + 2z + 2 = 0
   ∴ z = – 1 ± i
   R₁ = Res: (f(z): z = -1 + i)
   

   Ltz→-1±1[z-(-1+i)]	eix
   	[z-(-1+1)][z-(-1-i)]

   
   

   =	ei(-1+i)	=	e-i-1
   	-1 + i + 1 + i		2i

   
   
   

   = IM[πe-1(cos(1) - isin(1))] = -	π sin(1)
   	e

---

---

15. A parametric curve defined by x = cos		π		; y = sin		πu
    		2				2

    
    in the range 0 ≤ u ≤ 1 is rotated about the X-axis by 360 degrees. Area of the surface generated is

1. 1. π/2

   
   
   

   2. π

   
   
   

   3. 2π

   
   
   

   4. 4π

   
   
   

---

1. View Hint View Answer Discuss in Forum

   x = cos	πu	; y = sin		-πu
   	2			2

   
   x² + y² =1
   It represents a circle in x-y plane
   0 ≤ 4 ≤ 1
   ∴ 0 ≤ 4 ≤ 1, 0 ≤ y ≤ 1
   So, 0 ≤ θ < π/2
   Area of hemisphere = 2π(r)²
   = 2π(1)²
   = 2π

   Correct Option: C

   x = cos	πu	; y = sin		-πu
   	2			2

   
   x² + y² =1
   It represents a circle in x-y plane
   0 ≤ 4 ≤ 1
   ∴ 0 ≤ 4 ≤ 1, 0 ≤ y ≤ 1
   So, 0 ≤ θ < π/2
   Area of hemisphere = 2π(r)²
   = 2π(1)²
   = 2π

---