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Q.
The area enclosed between the curve $y = \log_e (x + e)$ and the coordinate axes is
Application of Integrals
Solution:
The graph of the curve $y = \log_e (x + e)$ is as shown in the fig.
Required area $A = \int\limits^{0}_{1-e} ydx = \int\limits^{0}_{1-e} \log_{e} \left(x +e\right)dx $
put $x + e = t \, \Rightarrow \, dx = dt $ also At $x = 1 - e, t = 1$
At $x = 0, t = e$ $\therefore \, A = \int\limits^{e}_1 \log_e tdt = [ t \log_e t - t]^{e}_{1}$
e - e - 0 +1 = 1
Hence the required area is 1 square unit.
