Question

If [t] denotes the greatest integer ≤ t, then the value of \(\int\limits_{0}^{1}\)[2x−|\(3x^2\)−5x+2|+1]dx is

• A. \(\frac{\sqrt37+\sqrt13-4}{6}\)
• B. \(\frac{\sqrt37-\sqrt13-4}{6}\)
• C. \(\frac{\sqrt-37-\sqrt13+4}{6}\)
• D. \(\frac{-\sqrt37+\sqrt13+4}{6}\)

Answer 1

The Correct Option is A

I=\(\int\limits_{0}^{1}\)[2x−|\(3x^2\)–5x+2|+1]dx

I=\(\int\limits_{0}^{2/3}\)\([\underbrace{ -3x^2+7x-2 }_{I1}]\)dx+\(\int\limits_{2/3}^{1}\)[\([\underbrace{ 3x^2-3x+2 }_{I2}]\)

\(I_1\)=\(\int\limits_{0}^{t_1}\)(−2)dx+\(\int\limits_{t_1}^{1/3}\)(−1)dx+\(\int\limits_{1/3}^{t_2}\)0.dx+\(\int\limits_{t_2}^{2/3}\)dx
=\(-t_1-t_2\)+\(\frac{1}{3}\),

Where, \(t_1=\frac{7-\sqrt37}{6}\), and \(t_2 = \frac{7-\sqrt13}{6}\)

\(I_2\)=\(\int\limits_{2/3}^{1}1dx=\frac{1}{3}\)
∴ \(I = \frac{1}{3}-t_1-t_2+\frac{1}{3}+1\)
=\(\frac{5}{3} - [\frac{7-\sqrt37}{6}+\frac{7-\sqrt13}{6}]\)

\(=\frac{\sqrt37+\sqrt13-4}{6}\)
So, the correct option is (A): \(\frac{\sqrt37+\sqrt13-4}{6}\)