How Do You Calculate the Area Under a Curve Using Antiderivatives?

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SUMMARY

The discussion focuses on calculating the area under the curve defined by the function f(x) = sqrt{2x - x²} between x = 0 and x = 2 using antiderivatives. The correct approach involves completing the square and applying trigonometric substitution. The transformation of the function to the form sqrt{1 - (x - 1)²} is crucial for selecting the appropriate trigonometric substitution. Participants emphasize the importance of understanding the chain rule and antiderivatives in this context.

PREREQUISITES
  • Understanding of antiderivatives and their properties
  • Knowledge of the chain rule in calculus
  • Familiarity with completing the square technique
  • Basic concepts of trigonometric substitution
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  • Study the method of trigonometric substitution for integrals
  • Practice completing the square with various quadratic expressions
  • Review the properties and applications of antiderivatives
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Students learning calculus, particularly those tackling integration techniques, and anyone seeking to understand the application of antiderivatives in calculating areas under curves.

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Homework Statement


As part of a bigger problem, I am trying to find the area under the curve
f(x) = sqrt{2x - x2} between x = 0 and x = 2. To do this, I have to find the antiderivative of f(x)

Homework Equations


antiderivative
f(x) = axb
F(x) = a/(b+1) * xb+1

chain rule
f(x) = a(b(x))
f'(x) = b'(x) * a'(b(x))

The Attempt at a Solution


I took the derivative of the terms in the parenthesis, then what was outside the parenthesis, to get
2/3 * (x2 - x3/3)3/2
I don't think this is right, because you can use the chain rule to find the derivative, and it's not the original function.

NOTE- I'm not actually old enough and am not actual in a calculus course right now, but I figure that the homework forum would be the best place to put this. Also, don't give me answer, but rather gove me a direction to go in.
 
Last edited:
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The best approach, short of looking at a table of integrals, is trig substitution, which you might not have seen yet. Before doing the substitution you need to do some work first, by completing the square.

\sqrt{2x - x^2} = \sqrt{-(x^2 - 2x)} = \sqrt{-(x^2 - 2x + 1) + 1} = \sqrt{1 - (x - 1)^2}

The next step is picking the right trig substitution.
 

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