Question

If f : [0, ∞) → \(\R\) and g : [0, ∞) → [0, ∞) are continuous functions such that
\(\int^{x^3+x^2}_0f(t)dt=x^2\) and \(\int_0^{g(x)}t^2dt=9(x+1)^3\) for all x ∈ [0, ∞),
then the value of
f(2) + g(2) + 16 f(12)
is equal to _________. (Rounded off to two decimal places)

Answer 1

Correct Answer: 13.39 - 13.41

To solve this problem, we need to find \( f(2) + g(2) + 16f(12) \). We'll tackle each part systematically:

1. Finding \( f(x) \): We begin with the given condition: \(\int^{x^3+x^2}_0f(t)dt=x^2\). Differentiate both sides with respect to \( x \) using the Fundamental Theorem of Calculus:
   \(f(x^3 + x^2) \cdot (3x^2 + 2x) = 2x\) 
   This implies:
   \(\displaystyle f(x^3 + x^2) = \frac{2x}{3x^2 + 2x}\)
   At \( x=2 \), \( x^3+x^2 = 12\), hence:
   \(\displaystyle f(12) = \frac{2 \cdot 2}{3 \cdot 2^2 + 2 \cdot 2} = \frac{4}{12 + 4} = \frac{1}{4}\)
   Thus, \(f(12) = 0.25\). For \( f(2) \), using \( x=2 \) gives:
   \(\displaystyle f(6) = \frac{2 \cdot 2}{3 \cdot 2^2 + 2 \cdot 2} = \frac{4}{12 + 4} = \frac{1}{4}\)
2. Finding \( g(x) \): Start with the given condition:\(\int_0^{g(x)}t^2dt=9(x+1)^3\). Integrating \( t^2 \) gives \( \frac{g(x)^3}{3} = 9(x+1)^3 \). Solving for \( g(x) \), we have:
   \(\displaystyle g(x) = \sqrt[3]{27(x+1)^3} = 3(x+1)\)
   Therefore, \( g(2) = 3(2 + 1) = 9 \).
3. Compute the final expression: Now we have:
   \(f(2) + g(2) + 16f(12) = \frac{1}{4} + 9 + 16 \cdot \frac{1}{4}\)
   \(<=>0.25 + 9 + 4 = 13.25\)

Thus, the final answer is 13.25.