JEE Mains · Maths · STD 12 - 7.2 definite integral
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
- A \(\frac{41}{3}\)
- B \(\frac{33}{2}\)
- C \(\frac{27}{2}\)
- D \(\frac{5}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{33}{2}\)
Step-by-step Solution
Detailed explanation
\(\because f\left(x^2+1\right)=x^4+5 x^2+2\) \(\left\{\right.\)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, \(\int_{0}^{3}f(t)dt=\int_{0}^{3}(t^{2}+3t-2)dt\) \([\frac{t^{3}}{3}+\frac{3t^{2}}{2}-2t]_{0}^{3}\)…
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