Question
using interation, find the area of the region bounded by the triangle ragion, the euations of whose sides are y = 2x + 1, y = 3x + 1 and x = 4.
✓
Answer
To find area of triangular region bounded by
y = 2x + 1 ...(i)
y = 3x + 1 ...(ii)
y = 4 ...(iii)
Equation (i) represents a parabola with vertex points(0, 1) and $\Big(-\frac{1}{2}, 0\Big)$ at origin and axis as x-axis and equation (ii) represents points (0, 1) and $\Big(\frac{1}{3}, 0\Big)$ a line parallel to y-axis.
Solving quation (i) and (ii) gives points B(0, 1)
Solving quation (ii) and (iii) gives points C(4, 13)
Solving quation (i) and (iii) gives points B(4, 9)
Area of rectargle = (y1 - y2)x
This apprectximating from x = 0, x = 4
Required area = Redion ABCA
$=\int\limits_{0}^{4}(\text{y}_{1}-\text{y}_{2})\text{dx} $
$=\int\limits_{0}^{4}\Big[3(\text{x}+1)-(2\text{x}+1)\Big]\text{dx} $
$=\int\limits_{0}^{4}\text{x}\text{dx}$
$=\Big[\frac{\text{x}^{2}}{2}\Big]^{4}_{0}$
$=8\ \text{sq.}\ \text{units}$
y = 2x + 1 ...(i)
y = 3x + 1 ...(ii)
y = 4 ...(iii)
Equation (i) represents a parabola with vertex points(0, 1) and $\Big(-\frac{1}{2}, 0\Big)$ at origin and axis as x-axis and equation (ii) represents points (0, 1) and $\Big(\frac{1}{3}, 0\Big)$ a line parallel to y-axis.
Solving quation (i) and (ii) gives points B(0, 1)
Solving quation (ii) and (iii) gives points C(4, 13)
Solving quation (i) and (iii) gives points B(4, 9)
Area of rectargle = (y1 - y2)x
This apprectximating from x = 0, x = 4
Required area = Redion ABCA
$=\int\limits_{0}^{4}(\text{y}_{1}-\text{y}_{2})\text{dx} $
$=\int\limits_{0}^{4}\Big[3(\text{x}+1)-(2\text{x}+1)\Big]\text{dx} $
$=\int\limits_{0}^{4}\text{x}\text{dx}$
$=\Big[\frac{\text{x}^{2}}{2}\Big]^{4}_{0}$
$=8\ \text{sq.}\ \text{units}$
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