Engineering Mathematics-I

For m, n > 0:

B(m, n) = Γ(m)Γ(n) / Γ(m + n).

From the ellipse, y² = b²(1 − x²/a²).

V = π ∫₋ₐᵃ y² dx

= (πb²/a²) ∫₋ₐᵃ (a² − x²) dx

= (4/3)πab².

The absolute series is Σₙ₌₁∞ 1/(n√n). Since this p-series has p = 3/2 > 1, the series converges absolutely.

Power series about c: Σₙ₌₀∞ aₙ(x − c)ⁿ = a₀ + a₁(x − c) + a₂(x − c)² + ⋯, where aₙ and c are constants.

A function is periodic if f(x + T) = f(x) for every x, where T > 0 is a fixed period.

Parseval’s theorem states that if

then

Substitute x = 1 and y = 2:

(1² + 7·2)/(1² + 2²) = 15/5 = 3.

Therefore, the limit is 3.

Since corresponding partial derivatives are equal, ∇×q = (1 − 1)i + (1 − 1)j + (1 − 1)k = 0.

Integrating first with respect to x gives

∫₀³∫₁² xy(1 + x + y) dx dy = ∫₀³ [(23/6)y + (3/2)y²] dy

= 123/4.

For a closed surface S bounding volume V,

∯ₛ F·n̂ dS = ∭ᵥ (∇·F) dV,

where n̂ is the outward unit normal.

Let t = 3θ. Then dθ = dt/3, and the limits become 0 to π/2.

I = (1/3) ∫₀^(π/2) cos⁴t sin²(2t) dt

= (4/3) ∫₀^(π/2) sin²t cos⁶t dt.

Using ∫₀^(π/2) sinᵐt cosⁿt dt = (1/2)B((m + 1)/2, (n + 1)/2),

I = (2/3)B(3/2, 7/2) = (2/3)(5π/128) = 5π/192.

The nth term is aₙ = xⁿ/[n(n + 1)]. By the ratio test,

limₙ→∞ |aₙ₊₁/aₙ| = limₙ→∞ |x|·n/(n + 2) = |x|.

So the series converges for |x| < 1 and diverges for |x| > 1.

At x = 1, Σ 1/[n(n + 1)] = Σ(1/n − 1/(n + 1)) = 1, so it converges. At x = −1, it converges absolutely because Σ 1/[n(n + 1)] converges.

Interval of convergence: [−1, 1]. Radius of convergence: 1.

The interval is not printed in the question, so assume the standard half-range interval 0 < x < π. Write

f(x) = a₀/2 + Σₙ₌₁∞ aₙ cos(nx),

where a₀ = (2/π)∫₀^π (2x − 1) dx = 2π − 2. Hence a₀/2 = π − 1.

For n ≥ 1,

aₙ = (2/π)∫₀^π (2x − 1) cos(nx) dx = 4((−1)ⁿ − 1)/(πn²).

Thus aₙ = 0 for even n and aₙ = −8/(πn²) for odd n. Therefore,

2x − 1 = π − 1 − (8/π)(cos x + cos 3x/3² + cos 5x/5² + ⋯),

for 0 < x < π.

Step 1: Find the gradient.

∇u = (2x + 4yz)i + (2y + 4xz)j + 4xyk.

At (1, −2, 2), ∇u = −14i + 4j − 8k.

Step 2: Find the unit vector for the given direction.

v = 2i − 2j + k, and |v| = √(4 + 4 + 1) = 3. Therefore, v̂ = (2i − 2j + k)/3.

Step 3: Take the dot product.

Dᵥ̂u = ∇u·v̂ = [−14(2) + 4(−2) + (−8)(1)]/3 = −44/3.

Directional derivative: −44/3.

Assume a > 0 and x, y > 0. By the arithmetic-geometric mean inequality,

xy + a³/x + a³/y ≥ 3[(xy)(a³/x)(a³/y)]^(1/3) = 3a².

Equality holds when xy = a³/x = a³/y. This gives x = y = a.

Minimum value: 3a², attained at x = y = a. There is no maximum for x, y > 0 because the function is unbounded as x = y increases.

The region is 0 ≤ x ≤ y < ∞. Reversing the order gives

I = ∫₀∞ ∫₀ʸ (e⁻ʸ/y) dx dy.

The integrand does not depend on x, so

I = ∫₀∞ (e⁻ʸ/y)[x]₀ʸ dy = ∫₀∞ e⁻ʸ dy

= [−e⁻ʸ]₀∞ = 1.

Use y = t and x = t²/4, where 0 ≤ t ≤ 4. Then dx = (t/2)dt and dy = dt.

For F = (P, Q), P = x²y² = t⁶/16 and Q = y = t. Hence

∫ F·dr = ∫₀⁴ (P dx + Q dy)

= ∫₀⁴ (t⁷/32 + t) dt

= [t⁸/256 + t²/2]₀⁴ = 256 + 8 = 264.

For the astroid x = a cos³t and y = a sin³t,

dx/dt = −3a cos²t sin t, and dy/dt = 3a sin²t cos t.

For 0 ≤ t ≤ π/2, ds = 3a sin t cos t dt. The surface area of this half is

S₁ = 2π∫₀^(π/2) y ds

= 6πa²∫₀^(π/2) sin⁴t cos t dt

= 6πa²[sin⁵t/5]₀^(π/2) = 6πa²/5.

The other half contributes the same area. Therefore, total surface area is

S = 2S₁ = 12πa²/5.

Let h = x − π/2. Then x = π/2 + h, so

sin x = sin(π/2 + h) = cos h.

The Taylor series for cos h is

cos h = 1 − h²/2! + h⁴/4! − h⁶/6! + h⁸/8! − ⋯.

Substituting h = x − π/2 gives

sin x = 1 − (x − π/2)²/2! + (x − π/2)⁴/4! − (x − π/2)⁶/6! + (x − π/2)⁸/8! − ⋯.

For −1 < x < 1, take l = 1. The Fourier series is

f(x) = a₀/2 + Σₙ₌₁∞ [aₙ cos(nπx) + bₙ sin(nπx)].

First, a₀ = ∫₋₁¹ (x − x²) dx = −2/3, so a₀/2 = −1/3.

For n ≥ 1, the odd term x cos(nπx) integrates to zero. Hence

aₙ = −2∫₀¹ x² cos(nπx) dx = 4(−1)ⁿ⁺¹/(n²π²).

Similarly, x² sin(nπx) is odd, so

bₙ = 2∫₀¹ x sin(nπx) dx = 2(−1)ⁿ⁺¹/(nπ).

Substituting the coefficients,

x − x² = −1/3 + Σₙ₌₁∞ [4(−1)ⁿ⁺¹ cos(nπx)/(n²π²) + 2(−1)ⁿ⁺¹ sin(nπx)/(nπ)],

for −1 < x < 1.

Assume a, b, c > 0 and l, m, n are not all zero. Let the constraints be

g = ax² + by² + cz² − 1 = 0, and h = lx + my + nz = 0.

Using Lagrange multipliers, form

L = x² + y² + z² − λ(ax² + by² + cz² − 1) − μ(lx + my + nz).

The stationary equations are

2x = 2λax + μl, 2y = 2λby + μm, 2z = 2λcz + μn,

with both constraints. The plane lx + my + nz = 0 is two-dimensional. In orthonormal coordinates on this plane, the restricted quadratic form has positive eigenvalues r₁ and r₂. The constraint becomes r₁s² + r₂t² = 1, while u = s² + t². Hence the extreme values of u are 1/r₁ and 1/r₂.

Let S = l² + m² + n². The sum and product of r₁ and r₂ are

τ = r₁ + r₂ = [a(m² + n²) + b(l² + n²) + c(l² + m²)]/S,

Δ = r₁r₂ = [bc·l² + ac·m² + ab·n²]/S.

Therefore, the extreme values satisfy Δu² − τu + 1 = 0. Thus

uₘᵢₙ = [τ − √(τ² − 4Δ)]/(2Δ),

uₘₐₓ = [τ + √(τ² − 4Δ)]/(2Δ).

The smaller root is the minimum and the larger root is the maximum.

Let P = x² − xy³ and Q = y² − 2xy. The square is traversed counterclockwise. Green’s theorem states

∮ (P dx + Q dy) around C = ∬ᴿ (∂Q/∂x − ∂P/∂y) dA.

Now, ∂Q/∂x = −2y and ∂P/∂y = −3xy². Therefore,

∂Q/∂x − ∂P/∂y = 3xy² − 2y.

For R = [0, 2] × [0, 2],

∬ᴿ (3xy² − 2y) dx dy = ∫₀²∫₀² (3xy² − 2y) dx dy

= ∫₀² (6y² − 4y) dy = [2y³ − 2y²]₀² = 16 − 8 = 8.

The line integral along the four sides also equals 8 (side contributions: 8/3, −16/3, 40/3, and −8/3). Both sides are equal, so Green’s theorem is verified.