Part B: 10-Mark Questions
QB. 1: (i) By using integration by parts, solve ∫ (log x)2 dx
Sol:
I = ∫ (log x)2 dx
u = (log x)2 ⇒ du/dx = 2(log x)/x ⇒ du = (2 log x)/x dx
∫ dv = ∫ dx ⇒ v = x
Now, ∫ u dv = uv - ∫ v du
∫ (log x)2 dx = x(log x)2 - ∫ x · (2 log x)/x dx
∫ (log x)2 dx = x(log x)2 - 2 ∫ log x dx
Now, let u = log x ⇒ du/dx = 1/x ⇒ du = (1/x) dx
Let ∫ dv = ∫ dx ⇒ v = x
∫ (log x)2 dx = x(log x)2 - 2[x log x - ∫ x · (1/x) dx]
= x(log x)2 - 2[x log x - ∫ dx]
= x(log x)2 - 2[x log x - x]
∫ (log x)2 dx = x(log x)2 - 2x log x + 2x
(ii) Find ∫01 tan-1x dx
∫ u dv = uv - ∫ v du
u = tan-1x ∫ dv = ∫ dx
du/dx = 1 / (1+x2) v = x
du = (1 / (1+x2)) dx
∫01 tan-1x dx = [x tan-1x]01 - ∫01 x · (1 / (1+x2)) dx.
= (1 · tan-1(1) - 0 · tan-1(0)) - ∫01 x / (1+x2) dx
∫01 tan-1x dx = π/4 - ∫01 x / (1+x2) dx
Let t = 1+x2 ⇒ dt/dx = 2x ⇒ dt = 2x dx ⇒ x dx = dt/2
Limits: x | 0 | 1
t | 1 | 2
∫01 tan-1x dx = π/4 - 1/2 ∫12 (1/t) dt = π/4 - 1/2 [log t]12
= π/4 - 1/2 [log 2 - log 1]
∫01 tan-1x dx = π/4 - (1/2)log 2
QB. 2: (i) Evaluate ∫ eax sin bx dx by using integration by parts.
Sol Let u = eax ∫ dv = ∫ sin bx dx
du/dx = a eax v = -cos bx / b
du = a eax dx
∫ eax sin bx dx = -eax cos bx / b + ∫ (cos bx / b) · a eax dx.
= -eax cos bx / b + a/b ∫ eax cos bx dx
Let u = eax ∫ dv = ∫ cos bx dx
du = a eax dx v = sin bx / b
∫ eax sin bx dx = -eax cos bx / b + a/b [ (eax sin bx / b) - ∫ (sin bx / b) · a eax dx ]
∫ eax sin bx dx = -eax cos bx / b + a/b2 (eax sin bx) - (a2/b2) ∫ eax sin bx dx
Let I = ∫ eax sin bx dx
I = (a/b2)eax sin bx - (1/b)eax cos bx - (a2/b2) I.
I + (a2/b2)I = (a/b2)eax sin bx - (1/b)eax cos bx.
Multiply by 'b2' on both sides,
b2I + a2I = a eax sin bx - b eax cos bx
I(a2+b2) = eax [a sin bx - b cos bx]
I = eax / (a2+b2) [a sin bx - b cos bx]
∫ eax sin bx dx = eax / (a2+b2) [a sin bx - b cos bx].
(ii) Change the order of integration and hence evaluate ∫04a ∫x2/4a2√(ax) xy dy dx
Sol:
1. Identify Region:
The region is bounded by:
- y = x2/(4a) ⇒ x2 = 4ay (Parabola opening up)
- y = 2√(ax) ⇒ y2 = 4ax (Parabola opening right)
- x = 0 to x = 4a
Intersection: (x2/4a)2 = 4ax ⇒ x4/(16a2) = 4ax ⇒ x4 = 64a3x
x(x3 - 64a3) = 0 ⇒ x=0 or x=4a.
When x=4a, y = (4a)2/(4a) = 4a.
The intersection point is (4a, 4a).
2. Change Order of Integration:
Original: Vertical strips (dy dx)
New: Horizontal strips (dx dy)
- y limits: 0 to 4a
- x limits: From y^2 = 4ax (left) to x^2 = 4ay (right)
⇒ x = y2/(4a) to x = 2√(ay)
3. Evaluate New Integral:
I = ∫04a ∫y2/(4a)2√(ay) xy dx dy
Inner (dx):
∫y2/(4a)2√(ay) xy dx = y [x2/2]x=y2/(4a)x=2√(ay)
= y/2 [ (2√(ay))2 - (y2/(4a))2 ]
= y/2 [ 4ay - y4/(16a2) ]
= 2ay2 - y5/(32a2)
Outer (dy):
∫04a (2ay2 - y5/(32a2)) dy
= [ 2ay3/3 - y6/(6 · 32a2) ]04a
= [ 2a(4a)3/3 - (4a)6/(192a2) ] - 0
= [ 2a(64a3)/3 - 4096a6/(192a2) ]
= [ 128a4/3 - (4096/192)a4 ]
= [ 128a4/3 - (64/3)a4 ]
= (128a4 - 64a4) / 3 = 64a4/3
QB. 3: Establish a reduction formula and evaluate In = ∫ sinnx dx . Hence, find I0 and I1
Sol: In = ∫ sinnx dx = ∫ sinn-1x sin x dx
Using integration by parts: ∫ u dv = uv - ∫ v du
Let u = sinn-1x ∫ dv = ∫ sin x dx ⇒ v = -cos x
du = (n-1)sinn-2x (cos x) dx
In = (sinn-1x)(-cos x) - ∫ (-cos x) [(n-1)sinn-2x cos x dx]
In = -sinn-1x cos x + (n-1) ∫ cos2x sinn-2x dx
= -sinn-1x cos x + (n-1) ∫ (1-sin2x) sinn-2x dx
= -sinn-1x cos x + (n-1) ∫ sinn-2x dx - (n-1) ∫ sinnx dx
In = -sinn-1x cos x + (n-1)In-2 - (n-1)In
In + (n-1)In = -sinn-1x cos x + (n-1)In-2
n In = -sinn-1x cos x + (n-1)In-2
In = (-sinn-1x cos x) / n + (n-1)/n In-2
Hence, find I0 and I1:
I0 = ∫ sin0x dx = ∫ 1 dx = x
I1 = ∫ sin x dx = -cos x
QB. 4: (i) Find the area of the region R enclosed by the parabola y=x2 and the line y=x+2.
Sol:
1. Find Intersection Points:
y = x2 ... (1)
y = x+2 ... (2)
Set them equal: x2 = x+2 ⇒ x2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
Points are: (2, 4) and (-1, 1).
2. Set up Double Integral for Area:
Area A = ∫∫R dy dx
We integrate using vertical strips.
- x limits: -1 to 2
- y limits: x2 (bottom curve) to x+2 (top curve)
3. Evaluate Integral:
A = ∫-12 ∫x2x+2 dy dx
Inner (dy):
∫x2x+2 dy = [y]x2x+2 = (x+2) - (x2) = x + 2 - x2
Outer (dx):
A = ∫-12 (x + 2 - x2) dx
A = [ x2/2 + 2x - x3/3 ]-12
A = [ ( (2)2/2 + 2(2) - (2)3/3 ) ] - [ ( (-1)2/2 + 2(-1) - (-1)3/3 ) ]
A = [ (4/2 + 4 - 8/3) ] - [ (1/2 - 2 + 1/3) ]
A = [ (2 + 4 - 8/3) ] - [ (3/6 - 12/6 + 2/6) ]
A = [ (6 - 8/3) ] - [ -7/6 ]
A = [ (18/3 - 8/3) ] + 7/6
A = 10/3 + 7/6
A = 20/6 + 7/6 = 27/6 = 9/2
The area is 9/2 square units.
(ii) Evaluate ∫03 ∫02 ex+y dy dx
I = ∫03 ∫02 ex+y dy dx
I = ∫03 ∫02 ex · ey dy dx = ∫03 ex [ey]02 dx
I = ∫03 ex [e2 - e0] dx = ∫03 ex (e2-1) dx
I = (e2-1) [ex]03 = (e2-1) (e3-e0)
I = (e2-1)(e3-1)
∫03 ∫02 ex+y dy dx = (e2-1)(e3-1).
QB. 5: Evaluate ∫∫ xy(x+y)dxdy over the area between y=x and y=x2
Sol:
1. Find Region:
Intersection of y=x and y=x2:
x = x2 ⇒ x2 - x = 0 ⇒ x(x-1) = 0
x = 0 or x = 1. Points are (0,0) and (1,1).
Region R: 0 ≤ x ≤ 1, and x2 ≤ y ≤ x
2. Set up Integral:
I = ∫01 ∫x2x xy(x+y) dy dx = ∫01 ∫x2x (x2y + xy2) dy dx
Inner (dy):
∫x2x (x2y + xy2) dy = [ x2y2/2 + xy3/3 ]y=x2y=x
= [ (x2(x)2/2 + x(x)3/3) ] - [ (x2(x2)2/2 + x(x2)3/3) ]
= [ x4/2 + x4/3 ] - [ x6/2 + x7/3 ]
= (5x4/6) - x6/2 - x7/3
Outer (dx):
I = ∫01 (5x4/6 - x6/2 - x7/3) dx
I = [ (5/6)(x5/5) - (1/2)(x7/7) - (1/3)(x8/8) ]01
I = [ x5/6 - x7/14 - x8/24 ]01
I = ( 1/6 - 1/14 - 1/24 ) - 0
LCM of 6, 14, 24 is 168.
I = ( 28/168 - 12/168 - 7/168 )
I = (28 - 19) / 168 = 9/168 = 3/56
QB. 6: (i) Using double integral, find the area bounded by y=x and y=x2
Sol:
1. Find Region:
Intersection of y=x and y=x2:
x = x2 ⇒ x2 - x = 0 ⇒ x(x-1) = 0
x = 0 or x = 1. Points are (0,0) and (1,1).
Region R: 0 ≤ x ≤ 1, and x2 ≤ y ≤ x
2. Set up Area Integral:
Area A = ∫∫R dy dx
A = ∫01 ∫x2x dy dx
Inner (dy):
∫x2x dy = [y]x2x = (x - x2)
Outer (dx):
A = ∫01 (x - x2) dx
A = [ x2/2 - x3/3 ]01
A = ( 1/2 - 1/3 ) - 0
A = 3/6 - 2/6 = 1/6
The required area = 1/6
(ii) Evaluate ∫05 ∫0x2 x(x2+y2) dy dx
Sol:
I = ∫05 ∫0x2 (x3+xy2) dy dx
Inner (dy):
∫0x2 (x3+xy2) dy = [ x3y + x(y3/3) ]0x2
= [ x3(x2) + x((x2)3/3) ] - 0
= x5 + x(x6/3)
= x5 + x7/3
Outer (dx):
I = ∫05 [x5 + x7/3] dx
I = [ x6/6 + (1/3)(x8/8) ]05
I = [ x6/6 + x8/24 ]05
I = ( (5)6/6 + (5)8/24 ) - 0
I = (5)6/6 + (5)8/24
QB. 7: (i) Evaluate ∫ t2et dt
Sol: I = ∫ t2et dt
Using Bernoulli's formula (tabular integration by parts):
∫ u dv = uv1 - u'v2 + u''v3 - ...
Let u = t2 ∫ dv = ∫ et dt
u' = 2t v1 = et
u'' = 2 v2 = et
u''' = 0 v3 = et
∫ t2et dt = (t2)(et) - (2t)(et) + (2)(et) + C
∫ t2et dt = et(t2 - 2t + 2) + C
(ii) Evaluate ∫ sin-1x dx
Sol: I = ∫ sin-1x dx = ∫ (sin-1x · 1) dx
Using integration by parts: ∫ u dv = uv - ∫ v du
Let u = sin-1x ∫ dv = ∫ 1 dx
du = (1 / √(1-x2)) dx v = x
I = ∫ sin-1x dx = x sin-1x - ∫ x (1 / √(1-x2)) dx
I = x sin-1x - ∫ x / √(1-x2) dx
To solve the new integral, let t = 1-x2
dt = -2x dx ⇒ x dx = -dt/2
∫ x / √(1-x2) dx = ∫ 1/√t · (-dt/2)
= -1/2 ∫ t-1/2 dt
= -1/2 [t1/2 / (1/2)] + C
= -t1/2 + C = -√(1-x2) + C
Substitute back:
I = x sin-1x - ( -√(1-x2) ) + C
∫ sin-1x dx = x sin-1x + √(1-x2) + C
QB. 8: Find the volume of the tetrahedron bounded by the co-ordinate planes and x/a + y/b + z/c = 1.
Sol: The volume V is given by the triple integral over the region.
The plane is z = c(1 - x/a - y/b).
The region in the xy-plane (where z=0) is 1 - x/a - y/b = 0, or y = b(1 - x/a).
x limits: 0 to a
y limits: 0 to b(1 - x/a)
z limits: 0 to c(1 - x/a - y/b)
V = ∫0a ∫0b(1-x/a) ∫0c(1-x/a-y/b) dz dy dx
Inner (dz):
∫0c(1-x/a-y/b) dz = [z]0... = c(1 - x/a - y/b)
Middle (dy):
∫0b(1-x/a) c(1 - x/a - y/b) dy
= c ∫0b(1-x/a) [ (1 - x/a) - y/b ] dy
= c [ (1 - x/a)y - y2/(2b) ]0b(1-x/a)
Let K = b(1-x/a). The integral is from 0 to K.
= c [ (1 - x/a)K - K2/(2b) ]
= c [ (K/b) · K - K2/(2b) ]
= c [ K2/b - K2/(2b) ] = c [ K2/(2b) ]
= c [ (b(1-x/a))2 / (2b) ]
= c [ b2(1-x/a)2 / (2b) ] = (bc/2) (1 - x/a)2
Outer (dx):
V = ∫0a (bc/2) (1 - x/a)2 dx
V = (bc/2) ∫0a (1 - 2x/a + x2/a2) dx
V = (bc/2) [ x - 2x2/(2a) + x3/(3a2) ]0a
V = (bc/2) [ x - x2/a + x3/(3a2) ]0a
V = (bc/2) [ (a - a2/a + a3/(3a2)) - (0) ]
V = (bc/2) [ a - a + a/3 ]
V = (bc/2) [ a/3 ]
V = abc/6
Volume = abc/6 Cubic units
QB. 9 (i): Evaluate ∫02a ∫0x ∫yx (xyz) dz dy dx
Sol:
I = ∫02a ∫0x ∫yx xyz dz dy dx
Inner (dz):
∫yx xyz dz = xy [z2/2]yx = (xy/2) [x2 - y2]
= (1/2) (x3y - xy3)
Middle (dy):
I2 = ∫0x (1/2) (x3y - xy3) dy
= (1/2) [ x3(y2/2) - x(y4/4) ]0x
= (1/2) [ (x3(x2/2) - x(x4/4)) - 0 ]
= (1/2) [ x5/2 - x5/4 ]
= (1/2) [ 2x5/4 - x5/4 ] = (1/2) [ x5/4 ] = x5/8
Outer (dx):
I = ∫02a (x5/8) dx
= (1/8) [ x6/6 ]02a
= (1/48) [ (2a)6 - 0 ]
= (1/48) [ 64a6 ]
= 64a6/48 = 4a6/3
QB. 9 (ii): Find the volume bounded by the cylinder x2+y2=4 and the planes y+z=4 and z=0.
Sol: Volume V = ∫∫R z dA
The region R is the circle x2+y2=4 in the xy-plane.
The height z is given by y+z=4 ⇒ z = 4-y.
V = ∫∫R (4-y) dA
Convert to polar coordinates:
Region R: 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
y = r sinθ
dA = r dr dθ
V = ∫02π ∫02 (4 - r sinθ) r dr dθ
V = ∫02π ∫02 (4r - r2 sinθ) dr dθ
Inner (dr):
∫02 (4r - r2 sinθ) dr = [ 4r2/2 - (r3/3) sinθ ]02
= [ 2r2 - (r3/3) sinθ ]02
= [ (2(2)2 - (2)3/3 sinθ) - 0 ]
= 8 - (8/3)sinθ
Outer (dθ):
V = ∫02π (8 - (8/3)sinθ) dθ
V = [ 8θ - (8/3)(-cosθ) ]02π
V = [ 8θ + (8/3)cosθ ]02π
V = [ (8(2π) + (8/3)cos(2π)) ] - [ (8(0) + (8/3)cos(0)) ]
V = [ 16π + 8/3(1) ] - [ 0 + 8/3(1) ]
V = (16π + 8/3) - 8/3
V = 16π
Volume = 16π cubic units
QB. 10: Obtain the volume of sphere x2+y2+z2=a2 by using triple integral.
Sol: We find the volume of one octant (where x,y,z ≥ 0) and multiply by 8.
V = 8 × (Volume of first octant)
Limits for first octant:
z limits: 0 to √(a2-x2-y2)
y limits: 0 to √(a2-x2)
x limits: 0 to a
V = 8 ∫0a ∫0√(a2-x2) ∫0√(a2-x2-y2) dz dy dx
Inner (dz):
∫0√(a2-x2-y2) dz = [z]0... = √(a2-x2-y2)
Middle (dy):
I2 = ∫0√(a2-x2) √((a2-x2)-y2) dy
Let b2 = a2-x2. This is ∫0b √(b2-y2) dy.
This integral represents the area of a quarter circle of radius b.
Area = (1/4)πb2
I2 = (1/4)π(a2-x2)
Outer (dx):
V = 8 ∫0a (1/4)π(a2-x2) dx
V = 2π ∫0a (a2-x2) dx
V = 2π [ a2x - x3/3 ]0a
V = 2π [ (a2(a) - a3/3) - 0 ]
V = 2π [ a3 - a3/3 ]
V = 2π [ 2a3/3 ]
V = 4πa3/3
The volume of the sphere is (4/3)πa3.