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Find the area of the region bounded by the parabola y = 4x^2, the tangent line to this parabola at (1, 4), and the x-axis.

Amarjeet K. | Professional Engineer for Math and Science TuroringProfessional Engineer for Math and Scien...
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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 :)

Amarjeet,

I like your answer except for one thing.  You don't need to integrate to get the area of the triangle.  The triangle has a base of 1/2 and height of 4 and the area is simple to compute as 1/2 * 4 * 1/2 = 1.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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Y = 4 X^2

Y' = 8X

Y'( 1) = 8     / Slope of the tangent line

4                               4
Bounded area = ∫ 4 X^2 dx = 4/3 X^3l    - 1/2 ( 4 ) ( 1/2) =

256/ 3 - 1 = 253/3

Graph will show that area of triangle subtracted from the area under the curve.