Question

How do you evaluate the definite integral $\int{\dfrac{1+{{y}^{2}}}{y}dy}$ from $\left[ 1,2 \right]$ ?

Answer 1

We have been given to calculate the definite integral of a fraction expression. It has to be integrated with respect to $dy$. We shall divide both the terms in the numerator by the denominator first to simplify the expression into integrable format. Since we have to perform definite integration, we shall also apply the upper limit and lower limits of integration.

Complete step by step solution:
Given that we have to find a definite integral of $\int{\dfrac{1+{{y}^{2}}}{y}dy}$ within the interval $\left[ 1,2 \right]$.
From this statement, we understand that the upper limit of the definite integral is 2 and the lower limit of definite integral is 1.
Thus, the definite integral is given as
$\int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}$
First of all, we shall divide both the terms in the numerator of the fraction by the denominator, y. Doing so will simplify our expression and make it easy to integrate.
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}=\int\limits_{1}^{2}{\left( \dfrac{1}{y}+\dfrac{{{y}^{2}}}{y} \right)dy}$
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}=\int\limits_{1}^{2}{\left( \dfrac{1}{y}+y \right)dy}$
We know that $\int{{{x}^{n}}.dx=\dfrac{{{x}^{n+1}}}{n+1}}+C$ and $\int{\dfrac{1}{x}.dx=\ln x}+C$.
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}=\left. \left( \ln y+\dfrac{{{y}^{2}}}{2} \right) \right|_{1}^{2}$
Applying the upper and lower limits of integration, we get
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}=\ln 2+\dfrac{{{2}^{2}}}{2}-\left( \ln 1+\dfrac{{{1}^{2}}}{2} \right)$
Since, $\ln 2=0.693$ and $\ln 1=0$, thus putting these values, we get
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}=0.693+2-0-\dfrac{1}{2}$
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}=0.693+2-0.5$
$\Rightarrow \int\limits_{1}^{2}{\dfrac{1+{{y}^{2}}}{y}dy}=2.193$
Therefore, the definite integral $\int{\dfrac{1+{{y}^{2}}}{y}dy}$ from $\left[ 1,2 \right]$ is equal to 2.193.

Note: Definite integral of a function $f\left( y \right)$ is the area bound under the graph of function, $x=f\left( y \right)$and above the y-axis which is bound between two bounds as $y=a$ and $y=b$. Here, $a=$ 1 and $b=$2. The best thing about 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 y-axis when the function is being integrated with respect to $dy$.