Question ID: #924
Let $f$ be a polynomial function such that $f(x^{2}+1)=x^{4}+5x^{2}+2,$ for all $x\in\mathbb{R}.$ Then $\int_{0}^{3}f(x)dx$ is equal to
- (1) $\frac{41}{3}$
- (2) $\frac{33}{2}$
- (3) $\frac{27}{2}$
- (4) $\frac{5}{3}$
Solution:
We are given $f(x^2 + 1) = x^4 + 5x^2 + 2$.
Let $t = x^2 + 1$. Then $x^2 = t – 1$.
Substitute $x^2$ into the equation:
$$f(t) = (t-1)^2 + 5(t-1) + 2$$
$$f(t) = (t^2 – 2t + 1) + 5t – 5 + 2$$
$$f(t) = t^2 + 3t – 2$$
So the function is $f(x) = x^2 + 3x – 2$.
We need to evaluate the definite integral $\int_{0}^{3} f(x) dx$.
$$\int_{0}^{3} (x^2 + 3x – 2) dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} – 2x \right]_{0}^{3}$$
Substitute the upper limit $x=3$:
$$= \left( \frac{3^3}{3} + \frac{3(3)^2}{2} – 2(3) \right) – 0$$
$$= 9 + \frac{27}{2} – 6$$
$$= 3 + 13.5$$
$$= 16.5 = \frac{33}{2}$$
Ans. (2)
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