Question

The value of integral $\underset{0}{\overset{1}{\int }}[3x^2+1]dx$ is
(where $[.]$ denotes the greatest integer function)

• A. $3-\sqrt{3}$
• B. $3+\frac{(\sqrt{2}-1)}{\sqrt{3}}$
• C. $3$
• D. $3-\frac{(\sqrt{2}+1)}{\sqrt{3}}$

Answer 1

The correct option is D $3-\frac{(\sqrt{2}+1)}{\sqrt{3}}$

Let, $\begin{array}{cc}I& =\underset{0}{\overset{1}{\int }}[3x^2+1]dx\\ \Rightarrow I& =\underset{0}{\overset{1}{\int }}\left[3x^2\right]dx+\underset{0}{\overset{1}{\int }}1dx\\ & =\underset{0}{\overset{1/\sqrt{3}}{\int }}0dx+\underset{1/\sqrt{3}}{\overset{\sqrt{2}/\sqrt{3}}{\int }}1\cdot dx+\underset{\sqrt{2}/\sqrt{3}}{\overset{1}{\int }}2dx+{x|}_0^1\\ & =0+(\frac{\sqrt{2}}{\sqrt{3}}-\frac{1}{\sqrt{3}})+2(1-\frac{\sqrt{2}}{\sqrt{3}})+1\\ & =3-\frac{(\sqrt{2}+1)}{\sqrt{3}}\end{array}$