Question

How do you evaluate the definite integral $\int{2{{e}^{x}}dx}$ from $\left[ 0,1 \right]$.

Answer 1

Hint: In this problem we need to evaluate the definite integral in the given range. In the given integral we have an exponential function along with a constant in multiplication. We know that we can’t integrate the constant. So, we will first take out the constant from the integration part. Now we will apply the integration formula $\int{{{e}^{x}}dx}={{e}^{x}}+C$. In the problem we have the definite integral, so we will apply the given limits and remove the constant. After simplifying the obtained equation, then we will get our required result.

Complete step-by-step solution:
Given that, $\int{2{{e}^{x}}dx}$.
In the above equation we have $2$ which is in multiplication. We know that we can’t integrate constantly. So, we are taking out the constant $2$ from the above equation, then we will get
$\int{2{{e}^{x}}dx}=2\int{{{e}^{x}}dx}$
Applying the integration formula $\int{{{e}^{x}}dx}={{e}^{x}}+C$ in the above equation, then we will get
$\Rightarrow \int{2{{e}^{x}}dx}=2\left[ {{e}^{x}}+C \right]$
In the problem they have asked to calculate the definite integral in the range of $\left[ 0,1 \right]$. So, applying the limits to the given integration, then we will get
$\Rightarrow \int\limits_{0}^{1}{2{{e}^{x}}dx}=2\left[ {{e}^{x}}+C \right]_{0}^{1}$
Applying the limits on the left side of the equation, then we will get
$\Rightarrow \int\limits_{0}^{1}{2{{e}^{x}}dx}=2\left( {{e}^{1}}-{{e}^{0}} \right)$
We know that ${{e}^{0}}=1$, then the above equation is modified as
$\Rightarrow \int\limits_{0}^{1}{2{{e}^{x}}dx}=2\left( e-1 \right)$

Note: In this problem they have asked to calculate the definite integral. So, we have first calculated the indefinite integral and then we have applied the limits. But here we have to remove the integration constant while applying the limits. So many times, we forget about the constant. But in integrals the integration constant has its unique importance.