Let f be a polynomial function such that f(x² + 1) = x⁴ + 5x² + 2

Let f be a polynomial function such that f(x² + 1) = x⁴ + 5x² + 2

Q. Let f be a polynomial function such that f(x² + 1) = x⁴ + 5x² + 2, for all x ∈ ℝ.

Then ∫₀³ f(x) dx is equal to

(A) 33/2

(B) 5/3

(C) 27/2

(D) 41/3

Correct Answer: 33/2

Explanation

Given,

f(x² + 1) = x⁴ + 5x² + 2

Let,

x² + 1 = t ⇒ x² = t − 1

Then,

f(t) = (t − 1)² + 5(t − 1) + 2

f(t) = t² − 2t + 1 + 5t − 5 + 2

f(t) = t² + 3t − 2

Hence,

f(x) = x² + 3x − 2

Now evaluate the integral:

∫₀³ (x² + 3x − 2) dx

= [ x³/3 + 3x²/2 − 2x ]₀³

= (27/3 + 27/2 − 6)

= 9 + 13.5 − 6

= 33/2

Hence, the required value is 33/2.

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