Question

Evaluate:${\int }_0^1{e}^{x}dx+{\int }_1^e\ln xdx$

Answer 1

${\int }_0^1{e}^{x}dx+{\int }_1^e\ln xdx$

Consider${\int }_1^e\ln xdx$

$={\int }_1^{e}1.\ln xdx$

Integrating by parts,using $\int u.dv=uv-\int v.du$

Let $u=\ln x\Rightarrow du=\frac{dx}{x}$

$dv=dx\Rightarrow v=x$

$={\left[x\ln x\right]}_1^e-{\int }_1^{e}x.\frac{dx}{x}$

$={\left[x\ln x\right]}_1^e-{\int }_1^{e}dx$

$={[x\ln x-x]}_1^e$

$=(e\ln e-e)-(\ln 1-1)$

$=(e\times 1-e)-(0-1)$

$=(e-e)+1$

$=1$

Now, ${\int }_0^1{e}^{x}dx+{\int }_1^e\ln xdx$

$={\left[e^x\right]}_0^1+1$(from above)

$=e^1-e^0+1$

$=e-1+1=e$