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Definite Integration: Continuous Function Satisfying

JEE Maths question with a full step-by-step solution.

Question
f:R→Rf:\mathbb R\to\mathbb R is a continuous function satisfying
2f(x)+f(x2+2)=x2+2x+3,2f(x)+f\left(\frac x2+2\right) = x^2+2x+3 ,
 then ∫03f(x) dx\displaystyle\int_0^3f(x)\,dx is
Solution
Answer: 6.33 (± 0.01)
Step 1: The map u↦u2+2u \mapsto \dfrac u2+2 sends [0,2]\left[0,2\right] onto [2,3]\left[2,3\right], so putting x=u2+2x = \dfrac u2+2, dx=du2dx = \dfrac{du}2,
∫23f(x) dx=12∫02f(u2+2)du\int_2^3f(x)\,dx = \frac12\int_0^2f\left(\frac u2+2\right)du
 Step 2: Splitting the required integral at x=2x = 2,
∫03f=∫02f(x) dx+∫23f(x) dx=∫02f(x) dx+12∫02f(x2+2)dx\int_0^3f = \int_0^2f(x)\,dx+\int_2^3f(x)\,dx = \int_0^2f(x)\,dx+\frac12\int_0^2f\left(\frac x2+2\right)dx
=12∫02[2f(x)+f(x2+2)]dx=12∫02(x2+2x+3)dx= \frac12\int_0^2\left[2f(x)+f\left(\frac x2+2\right)\right]dx = \frac12\int_0^2\left(x^2+2x+3\right)dx
 Step 3:
∫02(x2+2x+3)dx=83+4+6=383\int_0^2\left(x^2+2x+3\right)dx = \frac83+4+6 = \frac{38}3
∫03f(x) dx=12⋅383=193=6.3333…\int_0^3f(x)\,dx = \frac12\cdot\frac{38}3 = \frac{19}3 = 6.3333\ldots
 Step 4: Rounding to two decimals,
∫03f(x) dx=6.33\int_0^3f(x)\,dx = 6.33
 Answer: 6.336.33
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