intlimits_1^2 {{e^{2x}}} left( {frac{1}{x} - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx$. We know the standard integral form $\int e^{ax} [f(x) + \frac{f'(x)}{a}]

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx$. We know the standard integral form $\int e^{ax} [f(x) + \frac{f'(x)}{a}] dx = \frac{1}{a} e^{ax} f(x) + C$. Here,let $f(x) = \frac{1}{x}$. Then $f'(x) = -\frac{1}{x^2}$. Comparing with the given integral,$a = 2$. Thus,$\int e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx = \frac{1}{2} e^{2x} \left( \frac{1}{x} \right) = \frac{e^{2x}}{2x}$. Now,applying the limits from $1$ to $2$: $I = \left[ \frac{e^{2x}}{2x} \right]_1^2 = \frac{e^{2(2)}}{2(2)} - \frac{e^{2(1)}}{2(1)} = \frac{e^4}{4} - \frac{e^2}{2}$. This can be written as $\frac{e^4 - 2e^2}{4}$.

$\int\limits_1^2 {{e^{2x}}} \left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right)\,dx$ is equal to

Similar Questions

$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)\,dx} $ is equal to

$\int e^{x}\left(\frac{2}{x}-\frac{2}{x^2}\right) dx$ is equal to

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

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

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).

Vedclass Test Series

Exam Paper Generator

Online Exam Module