Definite Integration Question - None

Question

Let $f:[0,5] \to \mathbb{R}$ be such that $f''(x) = f''(5-x)$, $\forall x \in [0,5]$, $f'(0) = 1$ and $f'(5) = 7$, then the value of $\int_1^4 f'(x)\, dx$ is:

Accepted Answer

Step 1: Use the condition f''(x) = f''(5-x). Integrating both sides with respect to x:

f'(x) = f'(5-x) + C for some constant C

Step 2: Apply boundary conditions. At x = 0: f'(0) = f'(5) + C, so 1 = 7 + C, giving C = -6

Therefore: f'(x) = f'(5-x) - 6

Step 3: To find ∫₁⁴ f'(x)dx, use substitution. Let u = 5-x in half the integral:

∫₁⁴ f'(x)dx = ∫₁⁴ f'(x)dx

Note that for x ∈ [1,4], we have f'(x) + f'(5-x) = 2f'(5-x) - 6 + f'(5-x) = ... actually use direct pairing:

Step 4: Split: ∫₁⁴ f'(x)dx = ∫₁^2.5 f'(x)dx + ∫₂.₅⁴ f'(x)dx

For the second integral, substitute u = 5-x: ∫₂.₅⁴ f'(x)dx = ∫₁^2.5 f'(5-u)du = ∫₁^2.5 [f'(u) + 6]du

Therefore: ∫₁⁴ f'(x)dx = ∫₁^2.5 f'(x)dx + ∫₁^2.5 [f'(x) + 6]dx = ∫₁^2.5 [2f'(x) + 6]dx = 2∫₁^2.5 f'(x)dx + 6(1.5)

Step 5: By symmetry property: ∫₁⁴ f'(x)dx = [f(4) - f(1)] and using f'(x) + f'(5-x) = 6:

∫₁⁴ f'(x)dx = 6 × 3 = 18

Question:

Step-by-Step Solution

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