Area between a function and its tangent

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SUMMARY

The discussion focuses on calculating the area between the curve of the function f(x) = 4x², its tangent line at the point P(2, f(2)), and the x-axis. The tangent line's slope is determined to be 16, leading to the equation y = 16x - 30. Participants suggest integrating the area under the parabola and subtracting the area of the triangle formed by the tangent line and the x-axis as an alternative method. The correct approach involves using the integral of the difference between the functions and applying geometric area calculations.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and integrals.
  • Familiarity with the equation of a tangent line.
  • Knowledge of geometric area calculations, particularly for triangles.
  • Proficiency in manipulating polynomial functions.
NEXT STEPS
  • Study the process of finding the equation of a tangent line to a curve.
  • Learn how to compute definite integrals to find areas between curves.
  • Explore geometric area calculations, focusing on triangles and parabolic segments.
  • Investigate the Fundamental Theorem of Calculus for applications in area problems.
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Students studying calculus, particularly those focusing on integration and the geometric interpretation of derivatives, as well as educators seeking to enhance their teaching methods in these topics.

Momentum09
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Homework Statement


Find the area of the region bounded by the graph of f(x) = 4x^2, the tangent line to this graph at P(2, f(2)), and the x-axis


Homework Equations



Integral of [f(x)-g(x) dx]


The Attempt at a Solution


I first tried to find the equation for the tangent line
The derivative of 4x^2 = 8x, subsituting x=2 into the equation I got 16 [slope]
the equation then turned out to be y-2 = 16 (x-2) --> y = 16x-30
I then subtracted this tangent equation from 4x^2 then integrate, but I wasn't able to get the right answer.
Please show me how...thank you!
 
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Why did you have y - 2 = 16(x - 2)? If I plug in x = 2 I think it is not a tangent line...
 
Hi Momentum09! :smile:

I'll just add to what CompuChip :smile: says …

why not just integrate under the parabola in the usual way, and then subtract the area of the triangle using half-base-times-height? :wink:
 

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