Find are bounded by $y = e^{2 x} - 3 e^{x} + 2$

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Solution

$y = 0 ⟹ e^{x} = 1, 2 ⟹ x = 0, ln 2$

Area $= \int y d x = \int (e^{2 x} - 3 e^{x} + 2) d x$

$= [\frac{1}{2} e^{2 x} - 3 e^{x} + 2 x]_{0}^{ln 2}$

$= (\frac{1}{2} (4) - 3 (2) + 2 ln 2 - \frac{1}{2} + 3 + 0)$

$= 2 ln 2 - \frac{3}{2}$

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