Question

To find the value of the integral, \(\int_0^{10} | x - 5|dx\) by interpreting it in terms of its area.

Short answer

The value of the integral, \(\int_0^{10} | x - 5|dx\) is obtained as \(25\).

Step-by-step solution

Given data

The given integral is, \(\int_0^{10} | x - 5|dx\).

Concept used of Integral Calculus

We define integrals as the function of the area bounded by the curve\(y = f(x)\),\(a \le x \le b\), the\(x\)-axis, and the ordinates\(x = a\) and\(x = b\), where\(b > a\).

Let\(x\)be a given point in (\(a\),\(b\)). Then\(\int_a^b f (x)dx\)represents the area function.

This concept of area function leads to the fundamental theorems of integral calculus.

Represent integral as area

The given integral can be represented as the area of the region under the curve \(y = |x - 5|\) from 0 to 10 as follows.

Figure 1 shows the graph of the function, \(y = |x - 5|\) in the specified limit.

It is observed that there are two triangles formed namely 1 and 2 , above the \(x\)-axis in the graph.

Calculate the net area

The net area under the graph gives the value of the given integral.

Area of triangle 1 is obtained as follows.

\({A_1} = \frac{1}{2} \cdot 5 \cdot 5 = \frac{{25}}{2}\)

Similarly, obtain the area of triangle 2 .

\({A_2} = \frac{1}{2} \cdot 5 \cdot 5 = \frac{{25}}{2}\)

Therefore, the net area under the graph is computed as follows.

\(\begin{aligned}{c}A &= {A_1} + {A_2}\\ &= \frac{{25}}{2} + \frac{{25}}{2}\\ &= 25{\rm{ sq}}{\rm{. units}}\end{aligned}\)

Thus, the value of the integral, \(\int_0^{10} | x - 5|dx\) is obtained as \(\underline {25} \).