Question

The value of ${\int }_0^2{x}^{[x^2+1]}$ dx, where $\left[x\right]$ is the greatest integer less than or equal to $x$ is

• A. $\frac{469}{60}-\frac{1}{3}{2}^{3/2}-\frac{1}{5}{3}^{5/2}$
• B. $\frac{469}{60}+\frac{1}{3}{2}^{3/2}+\frac{1}{5}{3}^{5/2}$
• C. $\frac{469}{60}+\frac{1}{3}{2}^{3/2}-\frac{1}{5}{3}^{5/2}$
• D. $0$

Answer 1

The correct option is C $\frac{469}{60}+\frac{1}{3}{2}^{3/2}-\frac{1}{5}{3}^{5/2}$

In the interval $[0,1)$, $[x^2+1]=1$

In the interval $[1,\sqrt{2})$, $[x^2+1]=2$

In the interval $[\sqrt{2},\sqrt{3})$, $[x^2+1]=3$

In the interval $[\sqrt{3},2)$, $[x^2+1]=4$

Hence, ${\int }_0^2{x}^{[x^2+1]}dx$

$={\int }_0^1{x}^{1}dx+{\int }_1^{\sqrt{2}}{x}^{2}dx+{\int }_{\sqrt{2}}^{\sqrt{3}}{x}^{3}dx+{\int }_{\sqrt{3}}^2{x}^{4}dx$

$=\frac{1}{2}{\left[x^2\right]}_0^1+\frac{1}{3}{\left[x^3\right]}_1^{\sqrt{2}}+\frac{1}{4}{\left[x^4\right]}_{\sqrt{2}}^{\sqrt{3}}+\frac{1}{5}{\left[x^5\right]}_{\sqrt{3}}^2$

$=\frac{1}{2}+\frac{1}{3}(2^{3/2}-1)+\frac{1}{4}(9-4)+\frac{1}{5}(32-3^{5/2})$

$=\frac{469}{60}+\frac{1}{3}{2}^{3/2}-\frac{1}{5}{3}^{5/2}$

Hence, answer is option-(C).