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1 vote
Ex 3.6
6. find the area enclosed between the curve y= -2x²-5x+3 and the x-axis

User Charm
by
8.1k points

2 Answers

5 votes
y = -2x² - 5x + 3
-2x² - 5x + 3 = 0
x = -(-5) +/- √(-5² - 4(-2)(3))
2(-2)
x = 5 +/- √(25 + 24)
-4
x = 5 +/- √39
-4
x = 5 +/- 6.244997998398398
-4
x = 5 + 6.244997998398398 x = 5 - 6.244997998398398
-4 -4
x = 11.624997998398398 x = -1.244997998398398
-4 -4
x = -2.8112494995996 x = 0.3112494995996
-----------------------------------------------------------------------------------------------
y = -2x² - 5x + 3 y = -2x² - 5x + 3
y = -2(-3)² - 5(-3) + 3 y = -2(0.3)² - 5(0.3) + 3
y = -2(9) + 15 + 3 y = -2(0.09) - 0.15 + 3
y = -18 + 15 + 3 y = -0.18 - 0.15 + 3
y = -3 + 3 y = -0.33 + 3
y = 0 y = 2.67
-----------------------------------------------------------------------------------------------
(x, y) = (-3, 0) (x, y) = (0.3, 2.67)
User Ashikodi
by
8.4k points
6 votes
When y=0,


-2{ x }^( 2 )-5x+3=0\\ \\ 2{ x }^( 2 )+5x-3=0\\ \\ \left( 2x-1 \right) \left( x+3 \right) =0


\\ \\ \therefore \quad x=\frac { 1 }{ 2 } \\ \\ \therefore \quad x=-3

--------------------


\int _( -3 )^{ \frac { 1 }{ 2 } }{ -2{ x }^( 2 ) } -5x+3dx


\\ \\ ={ \left[ -\frac { { 2x }^( 2+1 ) }{ 2+1 } -\frac { 5{ x }^( 1+1 ) }{ 1+1 } +3x \right] }_( -3 )^{ \frac { 1 }{ 2 } }


\\ \\ ={ \left[ -\frac { 2{ x }^( 3 ) }{ 3 } -\frac { 5{ x }^( 2 ) }{ 2 } +3x \right] }_( -3 )^{ \frac { 1 }{ 2 } }


\\ \\ \\ =\left\{ -\frac { 2 }{ 3 } { \left( \frac { 1 }{ 2 } \right) }^( 3 )-\frac { 5 }{ 2 } { \left( \frac { 1 }{ 2 } \right) }^( 2 )+3\left( \frac { 1 }{ 2 } \right) \right\} -\left\{ -\frac { 2 }{ 3 } { \left( -3 \right) }^( 3 )-\frac { 5 }{ 2 } { \left( -3 \right) }^( 2 )+3\left( -3 \right) \right\}


\\ \\ \\ =-\frac { 2 }{ 3 } \cdot \frac { 1 }{ 8 } -\frac { 5 }{ 2 } \cdot \frac { 1 }{ 4 } +\frac { 3 }{ 2 } -\left\{ -\frac { 2 }{ 3 } \left( -27 \right) -\frac { 5 }{ 2 } \cdot 9-9 \right\}


\\ \\ =-\frac { 2 }{ 24 } -\frac { 5 }{ 8 } +\frac { 3 }{ 2 } -\left\{ \frac { 54 }{ 3 } -\frac { 45 }{ 2 } -9 \right\}


\\ \\ =-\frac { 2 }{ 24 } -\frac { 15 }{ 24 } +\frac { 36 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +9


\\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +\frac { 18 }{ 2 } \\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 63 }{ 2 }


\\ \\ =\frac { 343 }{ 24 }

Answer: 343/24 units squared.
User Wasit Shafi
by
8.6k points

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