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

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

Question

(A) We are given the integral $\int e^{x^2} \left( 2 - \frac{1}{x^2} \right) dx = e^{x^2} f(x) + C$. Let $I = \int e^{x^2} \left( 2 - \frac{1}{x^2} \r

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) We are given the integral $\int e^{x^2} \left( 2 - \frac{1}{x^2} \right) dx = e^{x^2} f(x) + C$. Let $I = \int e^{x^2} \left( 2 - \frac{1}{x^2} \right) dx$. We can rewrite the integrand as: $e^{x^2} \left( 2 - \frac{1}{x^2} \right) = e^{x^2} \left( 2x \cdot \frac{1}{x} - \frac{1}{x^2} \right)$. Notice that if we let $g(x) = \frac{1}{x}$,then $g'(x) = -\frac{1}{x^2}$. Thus,the integral becomes $\int e^{x^2} (2x g(x) + g'(x)) dx$. This is not quite in the standard form $\int e^{h(x)} (h'(x)g(x) + g'(x)) dx$. Let's differentiate $e^{x^2} f(x)$ to see if it matches the integrand: $\frac{d}{dx} [e^{x^2} f(x)] = e^{x^2} (2x) f(x) + e^{x^2} f'(x) = e^{x^2} [2x f(x) + f'(x)]$. Comparing this to $e^{x^2} (2 - \frac{1}{x^2})$,we get $2x f(x) + f'(x) = 2 - \frac{1}{x^2}$. If we assume $f(x) = \frac{1}{x}$,then $f'(x) = -\frac{1}{x^2}$. Substituting these into the equation: $2x(\frac{1}{x}) + (-\frac{1}{x^2}) = 2 - \frac{1}{x^2}$,which matches. Thus,$f(x) = \frac{1}{x} + k$. Given $f(\frac{1}{2}) = 2$,we have $\frac{1}{1/2} + k = 2 \implies 2 + k = 2 \implies k = 0$. Therefore,$f(x) = \frac{1}{x}$. Finally,$f(1) = \frac{1}{1} = 1$.

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).

Similar Questions

$\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x=f(x) e^{x}+C$,where $C$ is a constant,then $\frac{d^{3} f}{d x^{3}}$ at $x = 1$ is equal to

$\int {{e^{\sin x}}\left( {\sin x + {{\sec }^2}x} \right)} \,dx$ is equal to

$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x, x>0=$

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

$\int \frac{x e^x}{(1 + x)^2} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module