Don't definite Integrals find area?

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SUMMARY

The discussion centers on the confusion surrounding the calculation of the area under the curve defined by the function f(x) = 5 - x² over the interval [-2, 1]. The participant initially calculated the area using a Riemann Sum, arriving at a correct value of 12 square units. However, when applying the Fundamental Theorem of Calculus to compute the definite integral, they incorrectly evaluated the integral, leading to an erroneous result of -3. The correct evaluation of the definite integral should yield the same area as the Riemann Sum.

PREREQUISITES
  • Understanding of definite integrals and their properties
  • Familiarity with Riemann Sums and their application in calculating area
  • Knowledge of the Fundamental Theorem of Calculus
  • Basic algebra skills for evaluating polynomial functions
NEXT STEPS
  • Review the Fundamental Theorem of Calculus and its applications
  • Practice calculating areas using Riemann Sums for various functions
  • Learn how to correctly evaluate definite integrals with polynomial functions
  • Explore common mistakes in integral calculus and how to avoid them
USEFUL FOR

Students learning calculus, educators teaching integral calculus, and anyone seeking to clarify the relationship between Riemann Sums and definite integrals.

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I'm confused here.. My definite integral doesn't match by Riemman Sum... and it should right? I think that I have not integrated correctly. Can someone help me spot the problem? Thanks.

Find the Area of the region bounded by:
f(x)=5-x^2 , [-2, 1]

Using the Riemma Sum idea (limit of the sum of rectangles as the number of rectangles approaches infinity), I got 12 units^2 as my area, which is correct.

However, using definite integrals and the Fundamental Theorem of Calculus, I get:

\int_{-2}^{1} (5-x^2)dx}

=-\frac{x^3}{3}\biggl] ^{1}_{-2}

Which equals -3 ?
 
Physics news on Phys.org
5x - 1\3 * x^3
 
rocophysics said:
5x - 1\3 * x^3

Noooo.. I can't believe I made another stupid mistake! Thanks Roco
 

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