Question

Find the area bounded by the line y = x, the x – axis and the ordinates x = -1, x = 2.

Answer 1

Hint: We will first draw the graph corresponding to the situation given. Then we will use integration to integrate the function y = x between the limits x = -1, x = 2 to find the required area.

Complete step-by-step answer:

Now, before starting the solution we first to understand that the physical significance of $\int\limits_{a}^{b}{f\left( x \right)dx}$

Now, we can see from the graph that $\int\limits_{a}^{b}{f\left( x \right)dx}$ is nothing but the area bounded by the line $y=f\left( x \right),x=a,x=b\ and\ x-axis$.

Now, we have to find the area bounded by the line y = x, the x – axis and the ordinates x = -1, x = 2. So, the graph of this situation is,

Now, we know that integration finds the algebraic area under the curve that is if the curve is below the x – axis, then for that part the area will be negative. So, we have to find both of these areas separately to find the geometrical area.

So, we will write the bounded area as $ar\Delta ABO+ar\Delta OCD$.

So, we have the area of bounded region as,

$\left| \int\limits_{-1}^{0}{xdx} \right|+\left| \int\limits_{0}^{2}{xdx} \right|$

We have used different limits for both the areas as per the graph. Now, we know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$.

Also, we have used mod as the area is always positive. So, we have,

& =\left| \dfrac{{{x}^{2}}}{2} \right|_{-1}^{0}+\left| \dfrac{{{x}^{2}}}{2} \right|_{0}^{2} \\\ & =\left| \dfrac{{{0}^{2}}}{2}-\dfrac{{{\left( -1 \right)}^{2}}}{2} \right|+\left| \dfrac{{{2}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right| \\\ & =\left| \dfrac{1}{2} \right|+\left| 2 \right| \\\ & =2+\dfrac{1}{2} \\\ & =\dfrac{5}{2}sq\ units \\\ \end{aligned}$$ Therefore, the area bounded by the curve is $$\dfrac{5}{2}sq\ units$$. Note: It is important to notice that while finding the area of two separate regions we have used mod to tackle the case if the area comes to be negative as we know that the area of a region can’t be negative.