FYUG Even Semester Exam, 2024
MATHEMATICS (2nd Semester)
Integral Calculus and Vectors (MATDSC-152T)
Subject: MATHEMATICS
Paper Code: MATDSC-152T
Semester: 2nd Semester (FYUG)
Full Marks: 70
Time Duration: 3 Hours
SECTION-A
Answer any ten of the following questions (2 marks each).
Question 1
2
Express Integral(a to b) f(x) dx as the limit of sum.
Integral(a to b) f(x) dx = lim (h -> 0) h * [f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)]
where nh = b - a.
Question 2
2
Evaluate: Integral(-5 to 5) |x + 2| dx.
We split the integral at x = -2 where the absolute value changes sign:
= Integral(-5 to -2) -(x + 2) dx + Integral(-2 to 5) (x + 2) dx
= -[x^2/2 + 2x](-5 to -2) + [x^2/2 + 2x](-2 to 5)
= -[(2 - 4) - (12.5 - 10)] + [(12.5 + 10) - (2 - 4)]
= -[-2 - 2.5] + [22.5 + 2] = 4.5 + 24.5 = 29.
Question 3
2
Prove that Integral(0 to 2a) f(x) dx = Integral(0 to a) f(x) dx + Integral(0 to a) f(2a - x) dx.
We know Integral(0 to 2a) f(x) dx = Integral(0 to a) f(x) dx + Integral(a to 2a) f(x) dx.
In the second integral, let x = 2a - t, then dx = -dt.
When x=a, t=a; when x=2a, t=0.
Integral(a to 2a) f(x) dx = Integral(a to 0) f(2a-t) (-dt) = Integral(0 to a) f(2a-t) dt.
Replacing t by x, we get the required result.
Question 4
2
If In = Integral x^n cos(ax) dx and Jn = Integral x^n sin(ax) dx, show that a*In = x^n sin(ax) - n*Jn-1.
In = Integral x^n cos(ax) dx.
Using integration by parts (taking x^n as 1st function):
In = x^n * [sin(ax)/a] - Integral [n * x^(n-1) * (sin(ax)/a)] dx
In = (x^n sin(ax))/a - (n/a) * Integral x^(n-1) sin(ax) dx
Multiplying by 'a': a*In = x^n sin(ax) - n * Jn-1.
Question 5
2
Evaluate: Integral(0 to pi/2) cos^6(x) dx.
Using Reduction Formula: [(n-1)/n] * [(n-3)/(n-2)] * ... * (1/2) * (pi/2)
= (5/6) * (3/4) * (1/2) * (pi/2)
= 5*pi / 32.
Question 11
2
Find lambda if 2i-j+k, i+2j-k, and lambda*i-4j+5k are coplanar.
For coplanar vectors, the scalar triple product is zero:
| 2 -1 1 ; 1 2 -1 ; lambda -4 5 |
= 0
2(10 - 4) - (-1)(5 + lambda) + 1(-4 - 2*lambda) = 0
2(6) + 5 + lambda - 4 - 2*lambda = 0
12 + 1 - lambda = 0 => lambda = 13.
SECTION-B
Answer any five questions (10 marks each).
Question 16
5+5=10
(a) Evaluate: Limit(n -> infinity) [(1^2)/(n^3+1^3) + (2^2)/(n^3+2^3) + ... + (n^2)/(2n^3)].
The general term is r^2 / (n^3 + r^3) = (1/n) * (r/n)^2 / [1 + (r/n)^3].
Sum = Limit (1/n) * Sum[(r/n)^2 / (1 + (r/n)^3)] from r=1 to n.
This converts to Integral(0 to 1) [x^2 / (1 + x^3)] dx.
Let 1 + x^3 = t, then 3x^2 dx = dt.
= (1/3) Integral(1 to 2) (1/t) dt = (1/3) [log t](1 to 2) = (1/3) log 2.
(b) Prove that Integral(0 to pi/2) log(sin x) dx = -(pi/2) log 2.
Let I = Integral(0 to pi/2) log(sin x) dx. Also I = Integral(0 to pi/2) log(cos x) dx.
2I = Integral(0 to pi/2) [log(sin x) + log(cos x)] dx = Integral(0 to pi/2) log(sin x cos x) dx
2I = Integral(0 to pi/2) log(sin 2x / 2) dx = Integral(0 to pi/2) log(sin 2x) dx - (pi/2) log 2.
In first part, let 2x = u. Integral reduces to I.
2I = I - (pi/2) log 2 => I = -(pi/2) log 2.
Question 20
5+5=10
(a) Find the perimeter of the astroid x^(2/3) + y^(2/3) = a^(2/3).
Parametric equations: x = a*cos^3(t), y = a*sin^3(t).
dx/dt = -3a*cos^2(t)sin(t), dy/dt = 3a*sin^2(t)cos(t).
ds/dt = sqrt((dx/dt)^2 + (dy/dt)^2) = 3a*sin(t)cos(t).
Perimeter = 4 * Integral(0 to pi/2) 3a*sin(t)cos(t) dt
= 12a * [sin^2(t)/2](0 to pi/2) = 12a * (1/2) = 6a.
(b) Find the surface area of the solid generated by revolving the cardioid r = a(1 - cos theta) about initial line.
Surface Area S = Integral(0 to pi) 2*pi*y ds = Integral(0 to pi) 2*pi*r*sin(theta) * sqrt(r^2 + (dr/d*theta)^2) d*theta.
r = a(1-cos theta), dr/d*theta = a*sin theta.
sqrt(r^2 + (dr/d*theta)^2) = 2a*sin(theta/2).
After integration from 0 to pi, S = (32/5) * pi * a^2.
Question 21
5+5=10
(a) Prove the length of one arc of the cycloid x=a(theta-sin theta), y=a(1-cos theta) is 8a.
dx/d*theta = a(1-cos theta), dy/d*theta = a*sin theta.
ds/d*theta = sqrt[(dx/d*theta)^2 + (dy/d*theta)^2] = a*sqrt[(1-cos theta)^2 + sin^2 theta]
= a*sqrt[1 + cos^2 theta - 2cos theta + sin^2 theta] = a*sqrt[2 - 2cos theta]
= 2a*sin(theta/2).
Length = Integral(0 to 2*pi) 2a*sin(theta/2) d*theta = 2a * [-2cos(theta/2)](0 to 2*pi)
= -4a * [cos(pi) - cos(0)] = -4a * [-1 - 1] = 8a.
(b) Find the volume generated by revolution about x-axis of y^2 = x^2(2 - x).
Volume V = Integral(0 to 2) pi * y^2 dx
V = pi * Integral(0 to 2) x^2(2 - x) dx = pi * Integral(0 to 2) (2x^2 - x^3) dx
V = pi * [2x^3/3 - x^4/4](0 to 2) = pi * [16/3 - 16/4]
V = pi * [16/3 - 4] = pi * (4/3) = 4*pi/3 cubic units.
Would you like me to solve any other specific numerical from Section B in more detail?