Question

Using integration, find the area of the triangle, the equations of whose sides are
$y=2x+1,y=3x+1$and $x=4$

Answer 1

Hint: In this question, we first need to get the vertices of the triangle by solving the given three line equations. Then we get the limits within which we need to integrate. Now, on plotting the diagram we get the area by subtracting the areas formed by the two lines on integrating within the limits.

Complete step-by-step answer:
Now, from the given line equations in the question we get,

& y=2x+1 \\\ & y=3x+1 \\\ \end{aligned}$$ Now, on subtracting these two equations we get, $$\begin{aligned} & \Rightarrow -x=0 \\\ & \therefore x=0 \\\ \end{aligned}$$ Now, on substituting this value of x back in the line equation we get, $$\begin{aligned} & \Rightarrow y=2\times 0+1 \\\ & \therefore y=1 \\\ \end{aligned}$$ Now, we have one of the vertex as $$\therefore \left( x,y \right)=\left( 0,1 \right)$$ Let us now consider other two line equations $$\begin{aligned} & y=3x+1 \\\ & x=4 \\\ \end{aligned}$$ Here, as we already know the value of x on substituting this we get, $$\Rightarrow y=3\times 4+1$$ Now, on simplification we get, $$\Rightarrow y=13$$ $$\therefore \left( x,y \right)=\left( 4,13 \right)$$ Let us now consider another pair of lines $$\begin{aligned} & y=2x+1 \\\ & x=4 \\\ \end{aligned}$$ Now, on substituting the value of x we get, $$\Rightarrow y=2\times 4+1$$ Now, on further simplification we get, $$\Rightarrow y=9$$ $$\therefore \left( x,y \right)=\left( 4,9 \right)$$ Let us represent the vertices of the triangle as A, B, C $$A\left( 0,1 \right),B\left( 4,13 \right),C\left( 4,9 \right)$$ Now, let us plot the given lines  Here, we need to find the area of the shaded region As we already know from the applications of integration that The space occupied by the curve along with the axis, under the given condition is called area of bounded region. Area bounded by two curves $$y=f\left( x \right)$$ and $$y=g\left( x \right)$$ in which $$f\left( x \right)$$ lies above $$g\left( x \right)$$between $$x=a$$and $$x=b$$ is given by $$\Rightarrow \int\limits_{a}^{b}{\left\\{ f\left( x \right)-g\left( x \right) \right\\}dx}$$ Now, from the given conditions in the question we have $$\begin{aligned} & y=3x+1 \\\ & y=2x+1 \\\ \end{aligned}$$ Now, from the vertices obtained we have $$x=0,x=4$$ Now, on comparing these values with the formula above we have $$\begin{aligned} & f\left( x \right)=3x+1 \\\ & g\left( x \right)=2x+1 \\\ & a=0,b=4 \\\ \end{aligned}$$ Now, on substituting the respective values we get, $$\Rightarrow \int\limits_{a}^{b}{\left\\{ f\left( x \right)-g\left( x \right) \right\\}dx}$$ $$\Rightarrow \int\limits_{0}^{4}{\left\\{ 3x+1-\left( 2x+1 \right) \right\\}dx}$$ Now, on further simplification we get, $$\Rightarrow \int\limits_{0}^{4}{xdx}$$ As we already know from the properties of integration that $$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$$ Now, from the integration formula we get, $$\Rightarrow \left( \dfrac{{{x}^{2}}}{2} \right)_{0}^{4}$$ Now, on applying the limits we get, $$\Rightarrow \left( \dfrac{{{4}^{2}}}{2}-0 \right)$$ Now, on further simplification we get, $$\Rightarrow 8$$ Hence, the area of the triangle formed is 8 Note:Instead of integrating with respect to x in the given ordinates we can also calculate it by integrating it with respect to y from the values of the y obtained in the coordinates. Then integrate within the y coordinates to get the result. It is important to note that while calculating the coordinates we should not neglect any of the terms. It is also to be noted that we need to subtract the curve below from the curve above because considering it the other way changes the result completely.