Test: Solution Integrals - Engineering Mathematics MCQ

4x + 8 ≥ x² + 8

∴ x² – 4x ≤ 0

x(x – 4) ≤ 0 → (1)

Clearly the integral solution of (1) are 0, 1, 2, 3 and 4

∴ Total 5 values of x satisfies (1)

Given counter c is the circle, |z| = 1

⇒  z = e^iθ ⇒ dz = ie^iθdθ

Now, for upper half of the circle, 0 ≤ θ ≤ π

Calculation:

z = x + iy ⇒ dz = dx + i dy

given line is x = 2y ⇒ dy/dx = 1/2

lets substitute y in terms of x

I = ∫ x + i x² (dx + i/2 dx) 

I = ∫ x dx + i/2 x dx + i x² dx - x²/2 dx

Concept:

Integral of a complex function f(z) is given

∫ f(z) dz = ∫ (udx -vdy) + i ∫ (vdx + udy)

Noting f(z) = u(x, y) + i v(x, y) and dz = dx + i dy;

Calculation:

Given Along the straight line joining the points (0, 0) and (3, 1);

The equation of straight line will be x = 3y

⇒ dx = 3 dy ⇒ dz = (3 + i) dy;

Along the line x = 3y, the complex number z will be

z = x + iy = 3y + iy = (3 + i) y

Substituting both in the integral,

 has its poles at z = -1

and contour |z| = 2 is a circle of radius 2, centre (0, 0)

So