Question

How do you evaluate the definite integral $\int{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}}$ from $\left[ 0,1 \right]$ ?

Answer 1

Hint: We have been given to calculate the definite integral of a fractional expression written in terms of an exponential function. It has to be integrated with respect to $dx$. Since we have to perform definite integration, thus after writing the integral of exponential function, ${{e}^{x}}$by using the necessary integration, we shall also apply the upper limit and lower limits of integration. Further we will calculate the final solution after putting the values of ${{e}^{0}}$ and ${{e}^{1}}$.

Complete answer:
Given that we have to find definite integral of $\int{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}}$ within the interval $\left[ 0,1 \right]$.
From this statement, we understand that the upper limit of the definite integral is 1 and the lower limit of the definite integral is 0.
Thus, the definite integral is given as
$\int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}}$
In order to integrate this expression, we shall make use of a substitution method.
Let $t=1+{{e}^{x}}$ …………………… equation (1)
Since differential of ${{e}^{x}}$ is ${{e}^{x}}$ and differential of a constant is equal to zero, thus differentiating both sides, we get
$dt={{e}^{x}}.dx$
Applying the substitution of variable-t and dt, we get
$\Rightarrow \int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}.dx}=\int\limits_{0}^{1}{\dfrac{dt}{t}}$
We know that the integral is, $\int{\dfrac{dt}{t}}=\ln t+C$.
Substituting this value, we get
$\Rightarrow \int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}.dx}=\left. \ln t \right|_{0}^{1}$
Now, we shall re-substitute the value of variable-t from equation (1).
$\Rightarrow \int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}.dx}=\left. \ln \left( 1+{{e}^{x}} \right) \right|_{0}^{1}$
Applying the upper and lower limits of integration, we get
$\Rightarrow \int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}.dx}=\ln \left( 1+{{e}^{1}} \right)-\ln \left( 1+{{e}^{0}} \right)$
$\Rightarrow \int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}.dx}=\ln \left( 1+e \right)-\ln \left( 1+1 \right)$
$\Rightarrow \int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}.dx}=\ln \left( 1+e \right)-\ln 2$
Here, we will use the property of logarithm, $\ln a-\ln b=\ln \dfrac{a}{b}$.
 $\Rightarrow \int\limits_{0}^{1}{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}.dx}=\ln \dfrac{\left( 1+e \right)}{2}$
Therefore, the definite integral $\int{\dfrac{{{e}^{x}}}{1+{{e}^{x}}}}$ from $\left[ 0,1 \right]$ is equal to $\ln \dfrac{\left( 1+e \right)}{2}$.

Note:
Definite integral of a function $f\left( x \right)$ is the area bound under the graph of function, $y=f\left( x \right)$and above the x-axis which is bound between two bounds as $x=a$ and $x=b$. Here, $a=$ 0 and $b=$1. One of the powers of integral calculus is that the one of the two boundaries of the area to be found is a curve and the other boundary is the x-axis when the function is being integrated with respect to $dx$.