Question

The value of ${\int }_0^2\left(x^2\right]\mathrm{dx}$ is equal to

• A. $5-\sqrt{2}-\sqrt{3}$
• B. $5+\sqrt{2}+\sqrt{3}$
• C. $5+\sqrt{2}-\sqrt{3}$
• D. $5-\sqrt{2}+\sqrt{3}$

Answer 1

The correct option is A $5-\sqrt{2}-\sqrt{3}$

We have,
${\int }_0^2\left(x^2\right]\mathrm{dx}={\int }_0^1\left(x^2\right]\mathrm{dx}+{\int }_1^{\sqrt{2}}\left(x^2\right]\mathrm{dx}+{\int }_{\sqrt{2}}^{\sqrt{3}}\left(x^2\right]\mathrm{dx}+{\int }_{\sqrt{3}}^2\left(x^2\right]\mathrm{dx}={\int }_0^{1}0\cdot \mathrm{dx}+{\int }_1^{\sqrt{2}}1\cdot \mathrm{dx}+{\int }_{\sqrt{2}}^{\sqrt{3}}2\cdot \mathrm{dx}+{\int }_{\sqrt{3}}^{2}3\cdot \mathrm{dx}=0+{\left(x\right]}_1^{\sqrt{2}}+2{\left(x\right]}_{\sqrt{2}}^{\sqrt{3}}+3{\left(x\right]}_{\sqrt{3}}^2=\left(\sqrt{2}-1\right)+2\left(\sqrt{3}-\sqrt{2}\right)+3\left(2-\sqrt{3}\right)=5-\sqrt{2}-\sqrt{3}$