Evaluate the definite integral int_{1}^{2}lef | vedclass

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(A) We need to evaluate the integral $I = \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$.Let $2x = t$,then $2 dx = dt$,which impl

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$\frac{e^{4}-2e^{2}}{4}$

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(A) We need to evaluate the integral $I = \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$.Let $2x = t$,then $2 dx = dt$,which implies $dx = \frac{dt}{2}$.When $x = 1$,$t = 2$. When $x = 2$,$t = 4$.Substituting these into the integral:$I = \int_{2}^{4} \left(\frac{1}{t/2} - \frac{1}{2(t/2)^2}\right) e^t \frac{dt}{2}$$I = \int_{2}^{4} \left(\frac{2}{t} - \frac{2}{t^2/2}\right) e^t \frac{dt}{2} = \int_{2}^{4} \left(\frac{2}{t} - \frac{4}{t^2}\right) e^t \frac{dt}{2}$$I = \int_{2}^{4} \left(\frac{1}{t} - \frac{2}{t^2}\right) e^t dt$.This does not fit the standard form $\int e^t(f(t) + f'(t)) dt$ directly. Let us re-evaluate the substitution.Let $f(t) = \frac{1}{t}$,then $f'(t) = -\frac{1}{t^2}$.$I = \int_{2}^{4} e^t \left(\frac{1}{t} - \frac{1}{2t^2}\right) dt$ is not correct. Let's use the original expression:$I = \int_{1}^{2} \frac{e^{2x}}{x} dx - \int_{1}^{2} \frac{e^{2x}}{2x^2} dx$.Using integration by parts on $\int_{1}^{2} \frac{e^{2x}}{x} dx$ with $u = \frac{1}{x}$ and $dv = e^{2x} dx$:$du = -\frac{1}{x^2} dx$ and $v = \frac{e^{2x}}{2}$.$\int_{1}^{2} \frac{e^{2x}}{x} dx = \left[ \frac{e^{2x}}{2x} \right]_{1}^{2} - \int_{1}^{2} \frac{e^{2x}}{2} \left(-\frac{1}{x^2}\right) dx = \left[ \frac{e^{2x}}{2x} \right]_{1}^{2} + \int_{1}^{2} \frac{e^{2x}}{2x^2} dx$.Therefore,$I = \left[ \frac{e^{2x}}{2x} \right]_{1}^{2} + \int_{1}^{2} \frac{e^{2x}}{2x^2} dx - \int_{1}^{2} \frac{e^{2x}}{2x^2} dx = \left[ \frac{e^{2x}}{2x} \right]_{1}^{2}$.$I = \frac{e^4}{4} - \frac{e^2}{2} = \frac{e^4 - 2e^2}{4}$.

Evaluate the definite integral $\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$.

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