Question

How do you evaluate the definite integral $\int{\left| x-5 \right|dx}$ from $\left[ 0,10 \right]$ ?

Answer 1

Hint: We have been given to calculate the definite integral of an absolute value function. It has to be integrated with respect to $dx$. We analyze the given absolute value function in the given interval and hence, open it with the suitable positive or negative sign. 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 definite integral of $\int{\left| x-5 \right|dx}$ within the interval $\left[ 0,10 \right]$.
From this statement, we understand that the upper limit of the definite integral is 10 and the lower limit of the definite integral is 0.
On analyzing the absolute value function, $\left| x-5 \right|$, we see that it can be broken down into two intervals within the given interval.
$\left| x-5 \right|=\left\{ \begin{align}
  & -\left( x-5 \right),\text{ }-\infty < x\le 5 \\
 & \left( x-5 \right),\text{ }5\le x < \infty \text{ } \\
\end{align} \right.\text{ }$
Here, we shall apply the property of definite integrals and break the given integral into separate integrals according to their defined intervals as $\left[ 0,5 \right]$ and $\left[ 5,10 \right]$.
Thus, the definite integral is given as
$\int\limits_{0}^{10}{\left| x-5 \right|dx}=\int\limits_{0}^{5}{-\left( x-5 \right).dx}+\int\limits_{5}^{10}{\left( x-5 \right).dx}$
We know that $\int{{{x}^{n}}.dx=\dfrac{{{x}^{n+1}}}{n+1}}+C$ and \[\int{1.dx=x}+C\].
$\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=\int\limits_{0}^{5}{5-x.dx}+\int\limits_{5}^{10}{x-5.dx}$
$\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=\left. 5x-\dfrac{{{x}^{2}}}{2} \right|_{0}^{5}+\left( \left. \dfrac{{{x}^{2}}}{2}-5x \right|_{5}^{10} \right)$
Applying the upper and lower limits of integration, we get
$\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=5\left( 5 \right)-\dfrac{{{5}^{2}}}{2}-\left( 5\left( 0 \right)-\dfrac{{{0}^{2}}}{2} \right)+\dfrac{{{10}^{2}}}{2}-5\left( 10 \right)-\left( \dfrac{{{5}^{2}}}{2}-5\left( 5 \right) \right)$
$\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=5\left( 5 \right)-\dfrac{{{5}^{2}}}{2}+\dfrac{{{10}^{2}}}{2}-5\left( 10 \right)-\dfrac{{{5}^{2}}}{2}+5\left( 5 \right)$
$\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=25-2.\dfrac{{{5}^{2}}}{2}+\dfrac{100}{2}-50+25$
\[\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=50-50-2.\dfrac{25}{2}+\dfrac{100}{2}\]
Simplifying right hand side of the equation further, we get
\[\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=\dfrac{100}{2}-\dfrac{50}{2}\]
\[\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=\dfrac{50}{2}\]
\[\Rightarrow \int\limits_{0}^{10}{\left| x-5 \right|dx}=25\]

Therefore, the definite integral $\int{\left| x-5 \right|dx}$ from $\left[ 0,10 \right]$ is equal to 25.

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=$10. 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 x-axis when the function is being integrated with respect to $dx$.