Find $\frac{\mathrm{d}}{\mathrm{d}x}(2x-3)\mathrm{e}^{2x}$. Hence evaluate \(\displaystyle\display

Question

Find $\frac{\mathrm{d}}{\mathrm{d}x}(2x-3)\mathrm{e}^{2x}$. Hence evaluate $\displaystyle\displaystyle \displaystyle\int_{0}^{\ln 3} 4x\mathrm{e}^{2x} \, \mathrm{d}x$.

Accepted Answer

$\frac{\mathrm d}{\mathrm dx}[(2x-3)e^{2x}]=4xe^{2x}-4e^{2x}$ Hence: $\int_0^{\ln3}4xe^{2x}\,\mathrm dx=18\ln3-8$

Integration - Definite Integration as the Reverse of Differentiation

Integration - Definite Integration as the Reverse of Differentiation

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