Detailed step-by-step solutions for four improper integral problems, demonstrating how to evaluate limits to determine if an integral converges or diverges.

Let’s solve each integral step by step. We’ll check if they converge (give a finite number) or diverge (go to infinity).

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Exercise 1(a):
∫₀^∞ [x / (x² + 1)] dx

This is an improper integral because the upper limit is ∞. So we replace ∞ with L and take the limit as L → ∞.

Step 1: Write as limit
= lim(L→∞) ∫₀ᴸ [x / (x² + 1)] dx

Step 2: Use substitution to integrate
Let u = x² + 1 → du = 2x dx → so x dx = du/2

When x = 0, u = 1
When x = L, u = L² + 1

So integral becomes:
∫₁^(L²+1) (1/u) * (du/2) = (1/2) ∫₁^(L²+1) (1/u) du = (1/2)[ln|u|] from 1 to L²+1
= (1/2)[ln(L² + 1) - ln(1)] = (1/2) ln(L² + 1) [since ln(1)=0]

Step 3: Take limit as L → ∞
lim(L→∞) (1/2) ln(L² + 1) = ∞ (because ln of something going to infinity also goes to infinity)

✔ So this integral diverges.

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Exercise 1(b):
∫₁² [1 / √(x - 1)] dx

There’s a problem at x = 1 — the denominator becomes zero, so it’s undefined there. That means we have to approach x=1 from the right (from above).

Step 1: Rewrite as limit
= lim(L→1⁺) ∫ᴸ² (x - 1)^(-1/2) dx

Step 2: Integrate
Antiderivative of (x - 1)^(-1/2) is:
[ (x - 1)^(1/2) ] / (1/2) = 2(x - 1)^(1/2)

So evaluate from L to 2:
= lim(L→1⁺) [ 2(2 - 1)^(1/2) - 2(L - 1)^(1/2) ]
= lim(L→1⁺) [ 2(1) - 2√(L - 1) ]
= 2 - 2*0 = 2 [since as L→1⁺, √(L-1)→0]

✔ So this integral converges to 2.

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Exercise 2(a):
∫₀^ x e^(-x²) dx

Again, infinite interval → use limit.

Step 1: Write as limit
= lim(L→∞) ∫₀ᴸ x e^(-x²) dx

Step 2: Substitution
Let u = -x² → du = -2x dx → so x dx = -du/2

When x = 0, u = 0
When x = L, u = -L²

Integral becomes:
∫₀^(-L²) e^u * (-du/2) = (-1/2) ∫₀^(-L²) e^u du
But flipping limits removes the negative:
= (1/2) ∫_{-L²}^0 e^u du = (1/2)[e^u] from -L² to 0
= (1/2)[e⁰ - e^(-L²)] = (1/2)[1 - e^(-L²)]

Step 3: Take limit as L → ∞
e^(-L²) → 0, so we get (1/2)(1 - 0) = 1/2

✔ So this integral converges to 1/2.

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Exercise 2(b):
∫₁² [1 / (x - 1)²] dx

Discontinuity at x = 1 → approach from right.

Step 1: Rewrite as limit
= lim(L→1⁺) ∫ᴸ² (x - 1)^(-2) dx

Step 2: Integrate
Antiderivative of (x - 1)^(-2) is:
[ (x - 1)^(-1) ] / (-1) = -1/(x - 1)

Evaluate from L to 2:
= lim(L→1⁺) [ -1/(2 - 1) - ( -1/(L - 1) ) ]
= lim(L→1⁺) [ -1 + 1/(L - 1) ]

As L → 1⁺, (L - 1) → 0⁺, so 1/(L - 1) → ∞

Thus, the whole expression → ∞

✔ So this integral diverges.

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Final Answer:

Exercise 1(a): Diverges
Exercise 1(b): Converges to 2
Exercise 2(a): Converges to 1/2
Exercise 2(b): Diverges