Question

The value of the given integral $I={\int }_{-1}^0\frac{x^2}{e}dx+{\int }_0^1{x}^2.dx$

• A. $\frac{e+1}{3}$
• B. $\frac{e-1}{3e}$
• C. $\frac{e+1}{3e}$
• D. $\frac{1}{3e}$

Answer 1

The correct option is C

$\frac{e+1}{3e}$

Explanation for correct option
Given integral is $I={\int }_{-1}^0\frac{x^2}{e}dx+{\int }_0^1{x}^2.dx$
$\Rightarrow I={\int }_{-1}^0\frac{x^2}{e}dx+{\int }_0^1{x}^2.dx$
$$\begin{gathered} =\frac{1}{e}.{\left[\frac{x^3}{3}\right]}_{-1}^0+{\left[\frac{x^3}{3}\right]}_0^1\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{\int }{\mathbf{x}}^{\mathbf{n}}\mathbf{.}\mathbf{d}\mathbf{x}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{]} \\ =\frac{1}{e}.{\left[\frac{0^3}{3}-\frac{{\left(-1\right)}^3}{3}\right]}_{}^{}+{\left[\frac{1^3}{3}-\frac{0^3}{3}\right]}_{}^{} \\ =\frac{1}{3e}+\frac{1}{3} \\ =\frac{e+1}{3e} \end{gathered}$$

Hence option(C) is correct