JEE Mains · Maths · STD 12 - 7.2 definite integral
If for all real triplets \((a, b, c), f(x)=a+b x+c x^{2}\) then \(\int \limits_{0}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x}\) is equal to
- A \(\frac{1}{2}\left\{f(1)+3 f\left(\frac{1}{2}\right)\right\}\)
- B \(2\left\{3 f(1)+2 f\left(\frac{1}{2}\right)\right\}\)
- C \(\frac{1}{6}\left\{f(0)+f(1)+4 f\left(\frac{1}{2}\right)\right\}\)
- D \(\frac{1}{3}\left\{f(0)+f\left(\frac{1}{2}\right)\right\}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{6}\left\{f(0)+f(1)+4 f\left(\frac{1}{2}\right)\right\}\)
Step-by-step Solution
Detailed explanation
\(f(x)=a+b x+c x^{2}\) \(\int\limits_{0}^{1} f(x) d x=\left[a x+\frac{b x^{2}}{2}+\frac{c x^{3}}{3}\right]_{0}^{1}\) \(=a+\frac{b}{2}+\frac{c}{3}=\frac{1}{6}[6 a+3 b+c]\) \(=\frac{1}{6}\left[f(0)+f(1)+4 f\left(\frac{1}{2}\right)\right]\)
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