Question
Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 2), (1, 5) and (3, 4).
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Answer
Correct fiqure
Equation of
$\text{AB is}:\text{y} =\frac{1}{2}(3 \text{x} + 7 )$
$\text{BC is:}\text{y} = \frac{1}{2}(11 - \text{x})$
$\text{AC is}:\text{y} =\frac{1}{2}(\text{x} + 5 )$
Required area $ =\frac{1}{2}\int\limits_{-1}^{1}(3 \text{x} + 7 )\text{dx} + \frac{1}{2}\int\limits_{1}^{3}(11 - \text{x})\text{dx} - \frac{1}{2}\int\limits_{-1}^{3}(\text{x} + 5 )\text{dx}$
$ = \bigg[\frac{1}{12}(3 \text{x} + 7 )^{2}\bigg]_{-1}^{1} - \frac{1}{4}\bigg[(11-\text{x})^{2}\bigg]_{1}^{3} - \frac{1}{4}\bigg[(\text{x} + 5)^{2}\bigg]_{-1}^{3}$
= 7 + 9 – 12 = 4 sq. units.
Equation of
$\text{AB is}:\text{y} =\frac{1}{2}(3 \text{x} + 7 )$
$\text{BC is:}\text{y} = \frac{1}{2}(11 - \text{x})$
$\text{AC is}:\text{y} =\frac{1}{2}(\text{x} + 5 )$
Required area $ =\frac{1}{2}\int\limits_{-1}^{1}(3 \text{x} + 7 )\text{dx} + \frac{1}{2}\int\limits_{1}^{3}(11 - \text{x})\text{dx} - \frac{1}{2}\int\limits_{-1}^{3}(\text{x} + 5 )\text{dx}$
$ = \bigg[\frac{1}{12}(3 \text{x} + 7 )^{2}\bigg]_{-1}^{1} - \frac{1}{4}\bigg[(11-\text{x})^{2}\bigg]_{1}^{3} - \frac{1}{4}\bigg[(\text{x} + 5)^{2}\bigg]_{-1}^{3}$
= 7 + 9 – 12 = 4 sq. units.
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