Question

$\green{\underbrace{ \purple{ \underline{ \large\underline{\sf \: \red{Question:-}}}}}}$


Using integration find the area of region bounded by the triangle whose vertices are (-1,1),(0,5) and (3,2).

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$\\ \huge \red{\boxed{ \text{Answer :-}}} \\ \color{orangered}\boxed{\rm\frac{15}{2} sq.unit}$


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Answer 1

$\huge{ \color{lime} \widetilde{ \colorbox{black}{Ansꋪ}}}$

$\sf{ \rm{Let \: we \: have \: the \: vertices \: of  \: △ABC \:  as \:  A(−1,1),B(0,5) and C(3,2).}}</p><p></p><p>$

$\sf \rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: Equation \: of \:  AB \:  is  \: y−1=}</p><p></p><p>$

$\sf \rm{ \large( \frac{5 - 1}{0 + 1} ) \: \: \: \: \: \: \: \: \:{ \small \: (x + 1)}}$

$\sf{ \rm{ \implies{y−1 \: =4x+4}}}</p><p></p><p>$

$\sf{ \rm{ \implies{y=4x+5....(i)}}}$

$\sf{ \rm{ And \: equation \: of \:  BC  \: is \:  y−5=}}$

$\large{( \frac{2 - 5}{3 - 0}) \: \: \: \: \: \: \: \: \: \: { \small{(x - 0)}}}$

$\sf \implies{y−5= \frac{ - 3}{3} (x)}$

y=5−x....(ii)

$\sf{Similarly, \: equation \: of  \: AC  \: is  \: y−1=}$

$(\frac{2 - 1}{3 + 1}) \: \: \: \: \:( x + 1)$

$\sf{y−1= \frac{1}{4} (x + 1)}</em></strong></p><p><strong><em>[tex] \sf{y−1= \frac{1}{4} (x + 1)}$

$\implies{4y=x+5....(iii)}$

$\sf{Area \: of \: shaded \: region =}$

${∫}^{0} _{ - 1}(y_{1} - y _{2})dx + {∫}^{3} _{0}( y_{1} - y_{2})dx$

$= {∫}^{0} _{ - 1} \: \: [4x + 5 - \frac{x + 5}{4} ] \: dx + {∫}^{3} _{0} \: \: \: [5 - x - \frac{ x + 5}{4} ]dx$

$= [ \frac{ {4x}^{2} }{2} + 5x - \frac{ {x}^{2} }{8} - \frac{5x}{4} { ]_{ - 1} }^{0} +$

$= [5x - \frac{ {x}^{2} }{2} - \frac{ {x}^{2} }{8} - \frac{5x}{4} {]}^{3} _{0}$

$= [0 - (4. \frac{1}{2} + 5( - 1) - \frac{1}{8} + \frac{5}{5})] +$

$[(15 - \frac{9}{2} - \frac{9}{8} - \frac{15}{4}) - 0]$

$= 18 + ( \frac{1 - 10 - 36 - 9 - 30}{8} )$

$= 18 + ( - \frac{84}{8} ) = 18 - \frac{21}{2} = \frac{15}{2} sq \: units \:$

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hope its help u

Answer 2

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