give answer as a fraction and no decimals please
Find area under parabola from x =0 to x=1
Find eqaution of the tangent
Find the area under tangent from its x intersept to x=1
Subtract area under tangent from area under parabola
Area under parabola
y = 4x^2
This parabola passes through (0,0)
Definite Integrate from x =0 to point of tangent x= 1
Integral y = (4x^3)/3
Definite Integrate from x =0 to point of tangent x= 1 = 4 x 13/3 – 0 = 4/3 units
Equation of tangent line
Slope of parabola at (1,4)
Differentiate y = 4x^2
dy/dx = 8x = slope
slope at x=1 would be 8 x1 = 8
equation of tangent
y = mx + b
y = 8x + b
since this line passes through (1,4)
4 = 8 x 1 + b
b = -4
tangent equation y= 8x – 4
x intercept at y=0
0= 8x -4
X=1/2
Area under tangent from x= ½ to x=1
Integrate y = 8x - 4
Integral y = (8x^2)/2 – 4x
Definite integral from x= ½ to x= 1
((8 x 1)/2 – 4 x 1) – ((8 x ½ 2/2) – 4x1/2)
0 – (-1) = 1 unit
Subtract area under tangent from area under parabola
4/3 – 1 = 1/3 units – Answer :)
Find eqaution of the tangent
Find the area under tangent from its x intersept to x=1
Subtract area under tangent from area under parabola
Area under parabola
y = 4x^2
This parabola passes through (0,0)
Definite Integrate from x =0 to point of tangent x= 1
Integral y = (4x^3)/3
Definite Integrate from x =0 to point of tangent x= 1 = 4 x 13/3 – 0 = 4/3 units
Equation of tangent line
Slope of parabola at (1,4)
Differentiate y = 4x^2
dy/dx = 8x = slope
slope at x=1 would be 8 x1 = 8
equation of tangent
y = mx + b
y = 8x + b
since this line passes through (1,4)
4 = 8 x 1 + b
b = -4
tangent equation y= 8x – 4
x intercept at y=0
0= 8x -4
X=1/2
Area under tangent from x= ½ to x=1
Integrate y = 8x - 4
Integral y = (8x^2)/2 – 4x
Definite integral from x= ½ to x= 1
((8 x 1)/2 – 4 x 1) – ((8 x ½ 2/2) – 4x1/2)
0 – (-1) = 1 unit
Subtract area under tangent from area under parabola
4/3 – 1 = 1/3 units – Answer :)
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