Question

Using integration, find the area of the triangular region bounded by-
$y = 2x + 1, y = 3x + 1 and x = 4.$

Answer 1

Hint: Geometrically, definite integration can be used to find the area under the given curve up to the x-axis. We will first find the coordinates of the vertices of the triangle and then find the enclosing area using the formula-
\mathop \smallint \nolimits_{{\text{x}} = {\text{a}}}^{{\text{x}} = {\text{b}}} {\text{f}}\left( {\text{x}} \right)d{\text{x}} = area

Complete step-by-step answer:

To find the area enclosed by these lines, we will first find the point of intersection of these lines. Let the points be A, B and C respectively.
For point A,
$y = 2x + 1$
$y = 3x + 1$
Subtracting the two equations we get,
$x = 0$
$y = 2(0) + 1$
$y = 1$
So, the coordinates of A are A(0, 1)
For point B,
$y = 2x + 1$
$x = 4$
Substituting the value of x in the first equation,
$y = 2(4) + 1 = 9$
So, the coordinates of B are B(4, 9)
For point C,
$y = 3x + 1$
$x = 4$
Substituting the value of x in the first equation,
$y = 3(4) + 1 = 13$
So, the coordinates of C are C(4, 13)

Now we have to find the area of the triangle ABC formed by the given lines. By observing the figure, the area of triangle can be calculated as-
Area = area under the line AC - area under the line AB under the limits $x = 0 and x = 4$.
= \mathop \smallint \nolimits_0^4 \left( {3x + 1} \right)dx - \mathop \smallint \nolimits_0^4 \left( {2x + 1} \right)dx
$= \left( {\dfrac{{3{x^2}}}{2} + x} \right)_0^4 - \left( {\dfrac{{2{x^2}}}{2} + x} \right)_0^4$
$= \left( {\dfrac{{48}}{2} + 4 - 0} \right) - \left( {\dfrac{{32}}{2} + 4 - 0} \right)$
$= 24 + 4 - 16 - 4 = 8\;$
Area of the triangular region is 8 sq units. This is the required area.

Note: In such types of questions, we may need to divide the figure into two or more regions as shown in the figure. Also, a diagram is necessary to solve the problem correctly. Alternatively, we can use y-axis as the reference and then find the area of the triangle.