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# How do you find the integral of ${e^{\left( { - \dfrac{1}{2}} \right).x}}$?

Last updated date: 12th Sep 2024
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Hint:In order to determine the integral of the above exponential function ,use the method of integration by substitution by substituting $- \dfrac{1}{2}(x)$ with the $u$. Find the derivative of $- \dfrac{1}{2}x = u$ with respect to x and put the value of $dx$ in the original integral . Use the rule of integration $\int {{e^x}dx} = {e^c} + C$ to obtain the required result.
Formula:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$
$\int {{e^x}dx} = {e^c} + C$
$\dfrac{d}{{dx}}(x) = 1$

Complete step by step solution:
We are given a exponential function ${e^{\left( { - \dfrac{1}{2}} \right).x}}$ , whose integral will be
$I = \int {{e^{\left( { - \dfrac{1}{2}} \right).x}}dx}$---(1)
In order to integrate the above integral we will be using integration by substitution method by substituting $- \dfrac{1}{2}(x)$ with the $u$
So let $- \dfrac{1}{2}x = u$.
Differentiating the above with respect to $x$ using the rule of derivative that the derivative of variable $x$ is equal to one i.e. $\dfrac{d}{{dx}}(x) = 1$, we get
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( { - \dfrac{1}{2}x} \right) \\ \dfrac{{du}}{{dx}} = - \dfrac{1}{2}\dfrac{d}{{dx}}\left( x \right) \\ \dfrac{{du}}{{dx}} = - \dfrac{1}{2} \\ dx = - 2du \\$
Putting the value of $dx$ in the original integral , we get
$I = \int {{e^u}\left( { - 2du} \right)} \\ I = - 2\int {{e^u}du} \\$
And as we know that the integral of $\int {{e^x}dx} = {e^c} + C$ where C is the constant of integration .
$I = - 2{e^u} + C$
Putting back the value of $u$
$I = \int {{e^{\left( { - \dfrac{1}{2}} \right).x}}dx} = - 2{e^{\left( { - \dfrac{x}{2}} \right)}} + C$
Therefore, the integral$\int {{e^{\left( { - \dfrac{1}{2}} \right).x}}dx}$ is equal to $- 2{e^{\left( { - \dfrac{x}{2}} \right)}} + C$ where C is the constant of integration.
2. Indefinite integral=Let $f(x)$ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
3.The symbol $\int {f(x)dx}$ is read as the indefinite integral of $f(x)$ with respect to x.
4.Don’t forget to place the constant of integration $C$.