If \( [\bullet] \) denotes the greatest integer function, then the values of \( \int_{0}^{1.6}\displaystyle [x^2]dx \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.2-\sqrt{2} \)
\( [x^2] \) changes value where \( x^2 \) reaches an integer: \( x^2=1 \) at x = 1 and \( x^2=2 \) at \( x=\sqrt{2}\approx 1.414 \). At the upper limit \( (1.6)^2=2.56\lt 3 \).
\( \displaystyle 0\le x\lt 1:\ [x^2]=0 \)
\( \displaystyle 1\le x\lt\sqrt{2}:\ [x^2]=1 \)
\( \displaystyle \sqrt{2}\le x\le 1.6:\ [x^2]=2 \)
\( \displaystyle \int_{0}^{1.6}\displaystyle[x^2]\,dx \)
\( \displaystyle =\int_{0}^{1}\displaystyle 0\,dx+\int_{1}^{\sqrt{2}}\displaystyle 1\,dx+\int_{\sqrt{2}}^{1.6}\displaystyle 2\,dx \)
\( \displaystyle =0+(\sqrt{2}-1)+2(1.6-\sqrt{2}) \)
\( \displaystyle =\sqrt{2}-1+3.2-2\sqrt{2} \)
\( \displaystyle =2.2-\sqrt{2} \)