Question

How do you evaluate the definite integral $\int {\dfrac{{dx}}{{4 - x}}}$ from $\left[ {0,2} \right]$

Answer 1

Hint:At first, we will write the definite integration in the form $\int\limits_0^2 {\dfrac{{dx}}{{4 - x}}}$ . Then we will take the denominator $4 - x$ as a single variable $u$ or we can write $4 - x = u$ . Then we will differentiate both sides of the equation $4 - x = u$ w.r.t. $x$ . We will convert every term of $x$ in $u$ and then we will integrate the term. As the terms $x$ changed in $u$ we also have to change the interval $\left[ {0,2} \right]$ .

Formula Used : $\int {\dfrac{{dy}}{y}} = \log y + c$ ; where $c$ is an arbitrary constant.
If $\int {\phi (y)dy = F(y) + c}$ where c is an arbitrary constant then $\int\limits_a^b \phi (y)dy = F(b) - F(a)$ ..

Complete step by step answer:
We have to integrate $\int {\dfrac{{dx}}{{4 - x}}}$ on the interval $\left[ {0,2} \right]$ .
At first, we will write this in the form $\int\limits_0^2 {\dfrac{{dx}}{{4 - x}}}$ .
Let us choose;
$4 - x = u$
Differentiating both sides w.r.t. $x$ we will get;
As we know the differentiation of a constant w.r.t. a variable is zero.
$\Rightarrow - 1 = \dfrac{{du}}{{dx}}$
Simplifying we get;
$\Rightarrow dx = - du$
Now we will change the interval.
In the endpoint $x = 2$ ;
From our assumption $4 - x = u$ we get;
$u = 2$ .
In the starting point $x = 0$ ;
From our assumption $4 - x = u$ we get;
$u = 4$
Now we will transform each term $x$ in the term of $u$ and get;
$\int\limits_0^2 {\dfrac{{dx}}{{4 - x}}} = \int\limits_4^2 { - \dfrac{{du}}{u}}$
We know that $\int {\dfrac{{dy}}{y}} = \log y + c$ ; where $c$ is an arbitrary constant.
Now we will apply this formula $\int { - \dfrac{{du}}{u}}$ .
As the integration is definite we do not need to keep the constant part.
Now from $\int\limits_4^2 { - \dfrac{{du}}{u}}$ we will get;
$\Rightarrow \int\limits_4^2 { - \dfrac{{du}}{u}} = [ - \log u]_4^2$
Now we will apply the formula; if $\int {\phi (y)dy = F(y) + c}$ ; where c is an arbitrary constant then $\int\limits_a^b \phi (y)dy = F(b) - F(a)$ and get;
$\Rightarrow \int\limits_4^2 { - \dfrac{{du}}{u} = - \log 2 - ( - \log 4)}$
Simplifying the above equation we get;
$\Rightarrow \int\limits_4^2 { - \dfrac{{du}}{u}} = - \log 2 + \log 4$
As we know that $\log a - \log b = \log \dfrac{a}{b}$ we will get;
$\Rightarrow \int\limits_4^2 { - \dfrac{{du}}{u}} = \log \dfrac{4}{2}$
Simplifying the above equation we get;
$\Rightarrow \int\limits_4^2 { - \dfrac{{du}}{u}} = \log 2$

So, the correct answer is $\int\limits_0^2 {\dfrac{{dx}}{{4 - x}}} = \log 2$.

Note: Let, $\int {\phi (y)dy = F(y) + c}$ . Now if we use this formula under definite integration, we get$\int\limits_a^b \phi (y)dy = F(b) + c - F(a) - c$ . We can observe that the constant part will be cut out. That’s why in the definite integration there is no need to write the constant part.