Question

The area bounded by the curve y = log_e x and x-axis and the straight line x = e is

(a) e sq. units

(b) 1 sq. units

(c) 1−$\frac{1}{e}$ sq. units

(d) 1+$\frac{1}{e}$ sq. units

Answer 1

(b) 1 sq. units



The point of intersection of the curve and the straight line is A(e, 1).
Therefore, the area of the required region ABC,
$$\begin{gathered} A={\int }_0^1\left(x_1-x_2\right)\mathit{d}y\left(\mathrm{where},x_1=e\mathrm{and}{x}_2=e^y\right) \\ ={\int }_0^1\left(e-e^y\right)\mathit{d}y \\ ={\left[ey-e^y\right]}_0^1 \\ =\left\{e\left(1\right)-e^{\left(1\right)}\right\}-\left\{e\left(0\right)-e^{\left(0\right)}\right\} \\ =e-e+1 \\ =1\mathrm{square}\mathrm{unit} \end{gathered}$$