Mathematics - Definite Integration Question with Solution | TestHub

Let

$I={\int }_1^2{x}^2{e}^{\left[x\right]+\left[x^3\right]}dx$

$\Rightarrow I={\int }_1^2{x}^2{e}^{\left[{\mathrm{x}}^3\right]+1}dx$

$\Rightarrow I=e{\int }_1^2{x}^2{e}^{\left[{\mathrm{x}}^3\right]}dx$

Put $x^3=t$

$\Rightarrow 3x^{2}dx=dt$

So,

$I=\frac{e}{3}{\int }_1^8{e}^{\left[t\right]}dt$

$\Rightarrow I=\frac{e}{3}\left\{{\int }_1^{2}edt+{\int }_2^3{e}^{2}dt+\dots \dots \dots +{\int }_7^8{e}^{7}dt\right\}$

$\Rightarrow I=\frac{e}{3}\left(e+e^2+\dots \dots ..+e^7\right)$

$\Rightarrow I=\frac{e^2}{3}\frac{\left(e^7-1\right)}{(e-1)}$

So,

$\frac{3(e-1{)}^2}{e}{\int }_1^2{x}^2{e}^{\left[x\right]+\left[x^3\right]}dx=\frac{3}{e}\left(e-1\right)\times \frac{e^2}{3}\frac{\left(e^7-1\right)}{(e-1)}$

$\Rightarrow \frac{3(e-1{)}^2}{e}{\int }_1^2{x}^2{e}^{\left[x\right]+\left[x^3\right]}dx=e^8-e$