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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.10
Evaluate the definite integrals in Exercises 198 to 205 using substitution.
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Question : 205 of 271
Marks: +1, -0
∫12(1x−12x2)e2x dx\int\limits_{1}^{2} \left( \frac{1}{x} - \frac{1}{2x^2} \right) e^{2x} \, dx
Solution:  
Let I=∫12e2x(1x−12x2)dxI = \int\limits_{1}^{2} e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx
Put 2x=t⇒2dx=dt2x = t \Rightarrow 2dx = dt
When x=1,t=2x = 1, t = 2 and when x=2,t=4x = 2, t = 4
∴I=12∫24et(2t−1×42t2)dt\therefore I = \frac{1}{2} \int\limits_{2}^{4} e^t \left( \frac{2}{t} - \frac{1 \times 4}{2t^2} \right) dt
=12∫24et(2t−2t2)dt= \frac{1}{2} \int\limits_{2}^{4} e^t \left( \frac{2}{t} - \frac{2}{t^2} \right) dt
=∫24et(1t−1t2)dt= \int\limits_{2}^{4} e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) dt
=∫24et(1t+ddt(1t))dt= \int\limits_{2}^{4} e^t \left( \frac{1}{t} + \frac{d}{dt} \left( \frac{1}{t} \right) \right) dt
=[et⋅1t]24= \left[ e^t \cdot \frac{1}{t} \right]_{2}^{4}
=14et−e22= \frac{1}{4} e^t - \frac{e^2}{2}
=e22(e22−1)= \frac{e^2}{2} \left( \frac{e^2}{2} - 1 \right) or e2(e2−2)4\frac{e^2 (e^2 - 2)}{4}
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