FunctionHard
Question
Let $f$ be a polynomial function such that
$f\left( x^{2} + 1 \right) = x^{4} + 5x^{2} + 2$, for all $x \in \mathbb{R}$.
Then $\int_{0}^{3}\mspace{2mu} f(x)dx$ is equal to
Options
A.$\frac{41}{3}$
B.$\frac{33}{2}$
C.$\frac{27}{2}$
D.$\frac{5}{3}$
Solution
$\because f\left( x^{2} + 1 \right) = x^{4} + 5x^{2} + 2$
$\left\{ \right.\ $ put $\left. \ x^{2} + 1 = t \right\}$
$${\Rightarrow f(t) = (t - 1)^{2} + 5(t - 1) + 2 }{\Rightarrow f(t) = t^{2} + 3t - 2 }$$Now, $\int_{0}^{3}\mspace{2mu} f(t)dt = \int_{0}^{3}\mspace{2mu}\left( t^{2} + 3t - 2 \right)dt$
$${\left\lbrack \frac{t^{3}}{3} + \frac{3t^{2}}{2} - 2t \right\rbrack_{0}^{3} }{\left\lbrack \frac{27}{3} + \frac{27}{2} - 6 \right\rbrack }{= \frac{33}{2}}$$
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