Question

Find the area under the curve y =2x+1 bounded by x-axis , y-axis and the line x= 2 using integration.

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:y = 2x + 1 - - - (1)$

$\rm :\longmapsto\:x = 2 - - - (2)$

$\rm :\longmapsto\:x = 0 - - - (3)$

$\rm :\longmapsto\:y = 0 - - - (4)$

Now, we have to find the area bounded between these 4 lines.

Let first sketch the line (1).

$\rm :\longmapsto\:y = 2x + 1 - - - (1)$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:y = 2 \times 0 + 1$

$\rm :\longmapsto\:y = 0 + 1$

$\rm :\longmapsto\:y = 1$

Substituting 'x = 2' in the given equation, we get

$\rm :\longmapsto\:y = 2 \times 2 + 1$

$\rm :\longmapsto\:y = 4 + 1$

$\rm :\longmapsto\:y = 5$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 2 & \sf 5 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 1) & (2 , 5)

➢ See the attachment graph.

Consider,

$\bf :\longmapsto\:x = 2 - - - (2)$

is a line parallel to y - axis passes through the point (2, 0).

So, from graph, we conclude that 0ABC be the required region bounded between the curves y = 2x + 1, x = 2, x - axis and y - axis.

So, Required area is

$\rm \:  =  \:  \:\displaystyle\int_0^2\rm ydx$

$\rm \:  =  \:  \:\displaystyle\int_0^2\rm (2x + 1)dx$

We know,

$\boxed{ \sf{ \:\displaystyle\int\rm {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c}}$

and

$\boxed{ \sf{ \:\displaystyle\int\rm kdx = kx + c}}$

So, using these results,

$\rm \:  =  \:  \:\bigg( {x}^{2} + x \bigg)_0^2$

$\rm \:  =  \:  \:( {2}^{2} + 2) - ( {0}^{2} - 0)$

$\rm \:  =  \:  \:6 \: square \: units$