Definite Integration Question - None

Question

Evaluate $\int_0^1 (tx + 1 - x)^n\, dx$, $n \in \mathbb{N}$ and $t$ is a parameter independent of $x$. Also show that $$\int_0^1 x^k (1-x)^{n-k}\, dx = \frac{1}{(n+1)\cdot {}^nC_k}.$$

Accepted Answer

Step 1: Simplify the integrand

Rewrite the expression inside the integral:

tx + 1 - x = 1 + x(t - 1)

So the integral becomes: ∫₀¹ [1 + x(t-1)]^n dx

Step 2: Apply substitution

Let u = 1 + x(t-1), so du = (t-1)dx, giving dx = du/(t-1)

When x = 0: u = 1

When x = 1: u = 1 + (t-1) = t

The integral becomes:

∫₁ᵗ u^n · du/(t-1) = 1/(t-1) ∫₁ᵗ u^n du

Step 3: Evaluate the standard integral

∫₁ᵗ u^n du = [u^(n+1)/(n+1)]₁ᵗ = t^(n+1)/(n+1) - 1/(n+1) = [t^(n+1) - 1]/(n+1)

Step 4: Combine results

∫₀¹ (tx + 1 - x)^n dx = 1/(t-1) · [t^(n+1) - 1]/(n+1) = (t^(n+1) - 1)/[(n+1)(t-1)]

Step 5: Verify the Beta function result

Expand [1 + x(t-1)]^n using binomial theorem:

[1 + x(t-1)]^n = Σₖ₌₀ⁿ (n choose k)(t-1)^k x^k

Integrating term-by-term:

∫₀¹ Σₖ₌₀ⁿ (n choose k)(t-1)^k x^k dx = Σₖ₌₀ⁿ (n choose k)(t-1)^k · 1/(k+1)

For the special case when t = 0:

∫₀¹ (1-x)^n dx = (-1)^n/(n+1), and setting t = 0 in our formula with the binomial expansion yields ∫₀¹ x^k(1-x)^(n-k) dx

Using the Beta function identity: ∫₀¹ x^k(1-x)^(n-k) dx = k!(n-k)!/(n+1)! = 1/[(n+1)·(n choose k)]

∴ Answer: (t^(n+1) - 1)/((n+1)(t-1))

Question:

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