Question

How do you use substitution to integrate $e^{-5x}$ respect to x ?

Answer 1

Hint: The given question requires us to integrate a function of x with respect to x. Integration gives us a family of curves. Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc. Integral is always found with respect to some variable, which in this case is x.

Complete step by step solution:
The given question requires us to integrate an exponential function $e^{-5x}$ in variable x.
So, we have to compute $\int e^{-5x}dx$
The power or exponent is linear but not x simply. So, we need to do substitution in order to find the integral of the exponential function.
We can assign a new variable to the power of the exponential function $e^{-5x}$ .
Let us assume $-5x=t$ .
Differentiating both sides of the equation, we get,
 $\Rightarrow -5\left(dx\right)=dt$
Shifting $-5$ from left side of the equation to right side of the equation to find the value of $\left(dx\right)$ , we get,
 $\Rightarrow \left(dx\right)=\left({\displaystyle \frac{-dt}{5}}\right)$
So, the integral $\int e^{-5x}dx$ can be simplified by substituting the value of $\left(dx\right)$ as obtained above.
So, $\int e^{-5x}dx$
 $\Rightarrow \int e^t\left({\displaystyle \frac{-dt}{5}}\right)$
 $\Rightarrow \left({\displaystyle \frac{-1}{5}}\right)\int e^{t}dt$
We know that the integral of $e^x$ with respect to x is $e^x$ . So, we get,
 $\Rightarrow \left({\displaystyle \frac{-1}{5}}\right)e^t+C$
Putting the value of t back into the expression, we get,
 $\Rightarrow {\displaystyle \frac{-e^{-5x}}{5}}+C$
So, $\left({\displaystyle \frac{-e^{-5x}}{5}}+C\right)$ is the integral for the given function $e^{-5x}$ with respect to x.
So, the correct answer is “ $\left({\displaystyle \frac{-e^{-5x}}{5}}+C\right)$”.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant.