[tex]\green{\underbrace{ \purple{ \underline{ \large\underline{\sf \: \red{Question:-}}}}}} [/tex]
Using integration find the area of region bounded by the triangle whose vertices are (-1,1),(0,5) and (3,2).
[tex]\\ \rule{300pt}{0.1pt}[/tex]
[tex] \\ \huge \red{\boxed{ \text{Answer :-}}} \\ \color{orangered}\boxed{\rm\frac{15}{2} sq.unit}[/tex]
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Verified answer
[tex] \huge{ \color{lime} \widetilde{ \colorbox{black}{Ansꋪ}}}[/tex]
[tex] \sf{ \rm{Let \: we \: have \: the \: vertices \: of \: △ABC \: as \: A(−1,1),B(0,5) and C(3,2).}}
[/tex]
[tex] \sf \rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: Equation \: of \: AB \: is \: y−1=}
[/tex]
[tex] \sf \rm{ \large( \frac{5 - 1}{0 + 1} ) \: \: \: \: \: \: \: \: \:{ \small \: (x + 1)}}[/tex]
[tex] \sf{ \rm{ \implies{y−1 \: =4x+4}}}
[/tex]
[tex] \sf{ \rm{ \implies{y=4x+5....(i)}}}[/tex]
[tex] \sf{ \rm{ And \: equation \: of \: BC \: is \: y−5=}}[/tex]
[tex] \large{( \frac{2 - 5}{3 - 0}) \: \: \: \: \: \: \: \: \: \: { \small{(x - 0)}}}[/tex]
[tex] \sf \implies{y−5= \frac{ - 3}{3} (x)}[/tex]
y=5−x....(ii)
[tex] \sf{Similarly, \: equation \: of \: AC \: is \: y−1=}[/tex]
[tex] (\frac{2 - 1}{3 + 1}) \: \: \: \: \:( x + 1)[/tex]
[tex] \sf{y−1= \frac{1}{4} (x + 1)}
[tex] \sf{y−1= \frac{1}{4} (x + 1)}[/tex]
[tex] \implies{4y=x+5....(iii)}[/tex]
[tex] \sf{Area \: of \: shaded \: region =}[/tex]
[tex] {∫}^{0} _{ - 1}(y_{1} - y _{2})dx + {∫}^{3} _{0}( y_{1} - y_{2})dx[/tex]
[tex] = {∫}^{0} _{ - 1} \: \: [4x + 5 - \frac{x + 5}{4} ] \: dx + {∫}^{3} _{0} \: \: \: [5 - x - \frac{ x + 5}{4} ]dx[/tex]
[tex] = [ \frac{ {4x}^{2} }{2} + 5x - \frac{ {x}^{2} }{8} - \frac{5x}{4} { ]_{ - 1} }^{0} + [/tex]
[tex] = [5x - \frac{ {x}^{2} }{2} - \frac{ {x}^{2} }{8} - \frac{5x}{4} {]}^{3} _{0}[/tex]
[tex] = [0 - (4. \frac{1}{2} + 5( - 1) - \frac{1}{8} + \frac{5}{5})] + [/tex]
[tex] [(15 - \frac{9}{2} - \frac{9}{8} - \frac{15}{4}) - 0][/tex]
[tex] = 18 + ( \frac{1 - 10 - 36 - 9 - 30}{8} )[/tex]
[tex] = 18 + ( - \frac{84}{8} ) = 18 - \frac{21}{2} = \frac{15}{2} sq \: units \: [/tex]
[tex] \\ \rule{300pt}{0.1pt}[/tex]
hope its help u
Let we have the vertices of △ABC as A(−1,1),B(0,5) and C(3,2).
∴ Equation of AB is y−1=(
0+1
5−1
)(x+1)
⇒y−1=4x+4
⇒y=4x+5....(i)
And equation of BC is y−5=(
3−0
2−5
)(x−0)
⇒y−5=
3
−3
(x)
⇒y=5−x....(ii)
Similarly, equation of AC is y−1=(
3+1
2−1
)(x+1)
⇒y−1=
4
1
(x+1)
⇒4y=x+5....(iii)
∴ Area of shaded region =∫
−1
0
(y
1
−y
2
)dx+∫
0
3
(y
1
−y
2
)dx
=∫
−1
0
[4x+5−
4
x+5
]dx+∫
0
3
[5−x−
4
x+5
]dx
=[
2
4x
2
+5x−
8
x
2
−
4
5x
]
−1
0
+[5x−
2
x
2
−
8
x
2
−
4
5x
]
0
3
=[0−(4.
2
1
+5(−1)−
8
1
+
$
5
)]+[(15−
2
9
−
8
9
−
4
15
)−0]
=[−2+5+
8
1
−
4
5
+15−
2
9
−
8
9
−
4
15
]
=18+(
8
1−10−36−9−30
)
=18+(−
8
84
)=18−
2
21
= 15/2 square units