JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 5)
The area of the region enclosed by the curves $y=\mathrm{e}^x, y=\left|\mathrm{e}^x-1\right|$ and $y$-axis is :
$1+\log _{\mathrm{e}} 2$
$\log _{\mathrm{e}} 2$
$1-\log _{\mathrm{e}} 2$
$2 \log _{\mathrm{e}} 2-1$
Explanation
_24th_January_Evening_Shift_en_5_1.png)
$$\begin{aligned} &\text { For Area } \int_{-\ln 2}^0\left[e^x-\left(1-e^x\right)\right] d x\\ &\begin{aligned} & \int_{-\operatorname{nn} 2}^0\left(2 \mathrm{e}^{\mathrm{x}}-1\right) \mathrm{dx}=\left[2 \mathrm{e}^{\mathrm{x}}-\mathrm{x}\right]_{-\ln 2}^0 \\ & =(2-(1+\ell \mathrm{n} 2)) \\ & =1-\ell \mathrm{n} 2 \end{aligned} \end{aligned}$$
Comments (0)
