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November 26, 2014

November 26, 2014

Posted by **Ryan** on Wednesday, November 9, 2011 at 1:45pm.

- calculous -
**Steve**, Wednesday, November 9, 2011 at 8:27pmFirst, find the equation of the tangent line:

y = 4x^2

y' = 8x

slope at (4,64) = 32

(y-64)/(x-4) = 32

y = 32x - 64

32x-64 crosses the x-axis at x=2

So, we need to break the area up into two parts.

Area between the curve and y=0 on [0,2]

Area between curve and tangent line on [2,4]

Area = Int(4x^2 dx)[0,2] + Int(4x^2 - (32x-64))[2,4]

= (4/3 x^3)[0,2] + (4/3 x^3 - 16x^2 + 64x)[2,4]

= [4/3 * 8] + [4/3 * 64 - 16*16 + 64*4] - [4/3 * 8 - 16*4 + 64*2]

= 64/3

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