Engineering Mathematics Miscellaneous
- The integral ∮c(ydx - xdy) is evaluated along the circle x² + y² = 1/4 traversed in counter clockwise direction. The integral is equal to
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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 × RegionR
= – 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 × RegionR
= – 2 × area of circle with radius 1/2= 2 × π 1 ² = -π 2 2
- The area enclosed between the curves y² = 4x and x² = 4y is
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Given: y² = 4x
x² = 4y⇒ x4 = 4x 16
or x4 = 64x
or x(x³ – 64) = 0
or x³ = 64
or x = 4
∴ y = 4∴ Required area = 4∫0 √4x - x² dx y = 
2 × x3/2 × 2 - x³ 
4 3 12 0 = 4 × (4)x3/2 - 64 3 12 = 32 - 16 = 16 3 3 3
Alternately
For point where both parabolas cut each other
y² = 4x, x² = 4y
∴ x² = 4 √4x
or x² = 8√x
or x4 = 16 x
or x³ = 16
∴ x =(4, 0), (–4, 0)
∴ Required area= 4∫0√4x - 4∫0 x² dx = 2 × 2 x3/2 - x³ 4 = 16 4 3 12 0 3 Correct Option: A
Given: y² = 4x
x² = 4y⇒ x4 = 4x 16
or x4 = 64x
or x(x³ – 64) = 0
or x³ = 64
or x = 4
∴ y = 4∴ Required area = 4∫0 √4x - x² dx y = 2 × x3/2 × 2 - x³ 4 3 12 0 = 4 × (4)x3/2 - 64 3 12 = 32 - 16 = 16 3 3 3
Alternately
For point where both parabolas cut each other
y² = 4x, x² = 4y
∴ x² = 4 √4x
or x² = 8√x
or x4 = 16 x
or x³ = 16
∴ x =(4, 0), (–4, 0)
∴ Required area= 4∫0√4x - 4∫0 x² dx = 2 × 2 x3/2 - x³ 4 = 16 4 3 12 0 3
- The area enclosed between the straight line y = x and the parabola y = x² in the x–y plane is
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Given equations are y = x² ...(i)
y = x ...(ii)
From equations (i) and (ii)
x² – x = 0
→ x(x – 1) = 0
→ x = 0, 1
Area enclosed
= 1∫x=0y=x³∫y=xdy dx1∫x=0 dxy=x²∫y=x dy
= 1∫x=0[y]²x dx = 1∫x=0(x² - x)dx= x³ - x² 1 = 1 - 1 = 2 - 3 = - 1 3 2 0 3 2 6 6 ∴ Enclosed area = 
- 1 = 1 6 6 Correct Option: A
Given equations are y = x² ...(i)
y = x ...(ii)
From equations (i) and (ii)
x² – x = 0
→ x(x – 1) = 0
→ x = 0, 1
Area enclosed
= 1∫x=0y=x³∫y=xdy dx1∫x=0 dxy=x²∫y=x dy
= 1∫x=0[y]²x dx = 1∫x=0(x² - x)dx= x³ - x² 1 = 1 - 1 = 2 - 3 = - 1 3 2 0 3 2 6 6 ∴ Enclosed area = - 1 = 1 6 6
- The value of the definite integral e∫1 √x In (x)dx is
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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)
- The value of the integral
∞∫-∞ sin x dx is x² + 2x + 2
evaluated using contour integration and the residue theorem is
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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
