
How do you find the integral of \[{e^{\left( { - \dfrac{1}{2}} \right).x}}\]?
Answer
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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.
Additional Information:
1.Different types of methods of Integration:
Integration by Substitution
Integration by parts
Integration of rational algebraic function by using partial fraction
2. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.
Note:
1.Use standard formula carefully while evaluating the integrals.
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 $C$.
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.
Additional Information:
1.Different types of methods of Integration:
Integration by Substitution
Integration by parts
Integration of rational algebraic function by using partial fraction
2. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.
Note:
1.Use standard formula carefully while evaluating the integrals.
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 $C$.
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