Question

If $[t]$ denotes the greatest integer $\le t$, then the value of $\frac{3(e-1)}{e}\underset{1}{\overset{2}{\int }}{x}^2{e}^{[x]+\left[x^3\right]}dx$ is :

• A. $e^8-1$
• B. $e^7-1$
• C. $e^9-e$
• D. $e^8-e$

Answer 1

Answer: (D)

$\underset{1}{\overset{2}{\int }}{x}^2{e}^{\left[x^3\right]+1}dx$
$x^3=t$
$3x^{2}dx=dt$
$=\frac{e}{3}\underset{1}{\overset{8}{\int }}{e}^{[t]}dt$
$=\frac{e}{3}\left\{\underset{1}{\overset{2}{\int }}edt+{\int }_2^3{e}^{2}dt+\dots \dots ..+\underset{7}{\overset{8}{\int }}{e}^{7}dt\right\}$
$=\frac{e}{3}\left(e+e^2+\dots \dots \dots +e^7\right)$
$=\frac{e^2}{3}\left(1+e+\dots \dots \dots +e^6\right)=\frac{e^2}{3}\frac{\left(e^7-1\right)}{(e-1)}$
$\frac{3(e-1)}{e}\underset{1}{\overset{2}{\int }}{x}^2\times e^{[x]+\left[x^3\right]}dx=\frac{3}{e}(e-1)\times \frac{e^2}{3}\frac{\left(e^7-1\right)}{(e-1)}$
$=e\left(e^7-1\right)$
$=e^8-e$