Area of triangle by integrals (is this right?)

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SUMMARY

The area of a triangle with vertices (0,5), (-2,-2), and (2,2) can be accurately calculated using integrals, resulting in an area of A=10. The method involves evaluating the definite integral of the function derived from the triangle's vertices. The discussion confirms that the approach is valid, as verified by Heron's method. Users are encouraged to double-check their calculations to ensure accuracy.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with the concept of vertices in geometry
  • Knowledge of Heron's formula for area calculation
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the application of definite integrals in calculating areas of irregular shapes
  • Learn about Heron's formula for triangle area calculation
  • Explore the geometric interpretation of integrals
  • Practice solving integrals involving polynomial functions
USEFUL FOR

Students in calculus, geometry enthusiasts, and anyone looking to deepen their understanding of area calculations using integrals.

silicon_hobo
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[SOLVED] area of triangle by integrals (is this right?)

Homework Statement


Use integrals to find the area of a triangle with vertices (0,5),(-2,-2),(2,2).

Homework Equations



The Attempt at a Solution


I think I've got it. I'm just looking for some confirmation of my method before I move on.
Thanks for your time. Cheers.

http://www.mcp-server.com/~lush/shillmud/quest3.jpg

P.S. How do I add a [Solved] to the title?
 
Last edited:
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How are you getting e.g. integral of 5x/2+5 to be 5x^2/4+5x^2/2??
 
Ooops. Thanks for pointing that out Dick. I found the dx of 5x instead of the dx of 5 which interestingly seems not to have affected my answer. I guess you can see why I'm using the forum. Everything should be in order now. Cheers.
 
silicon_hobo said:
Ooops. Thanks for pointing that out Dick. I found the dx of 5x instead of the dx of 5 which interestingly seems not to have affected my answer. I guess you can see why I'm using the forum. Everything should be in order now. Cheers.

If you are still getting 20 for the area, I can't agree with that. Double check again.
 
Okay, I went through again and got A=10. When putting in values for x to solve I sub the rightmost value from the top of integral sign and then subtract from that the same antiderivative with the leftmost x value (top of the integral sign). This is correct? Thanks!
 
Yes, A=10 works.
 
Thanks Dick. Heron's method confirms it. Now how do I mark this one [SOLVED] ?
 
Under Thread Tools, at the top, isn't there a "Mark this thread as solved"?
 

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