Area under Curve: Find x-intercept & y-intercept

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SUMMARY

The discussion focuses on calculating the area bounded by the curve defined by the equation y = x - 2√x and the x-axis (y = 0). The x-intercept is determined to be 4, while the y-intercept is 0. The area calculation involves integrating the difference between the upper curve and the lower curve, leading to the conclusion that the area A is 16. A correction is noted, indicating that the correct integral to evaluate is ∫(2√x - x) from 0 to 4.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with the concept of x-intercepts and y-intercepts
  • Knowledge of area under a curve calculations
  • Experience with evaluating definite integrals
NEXT STEPS
  • Study the process of finding the area under curves using definite integrals
  • Learn about the properties of x-intercepts and y-intercepts in graphing functions
  • Explore the application of integration techniques in calculus, specifically for curves
  • Review the method of using vertical rectangles for area calculations
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Students studying calculus, particularly those focusing on integral calculus and area calculations, as well as educators looking for examples of area under curves and intercepts.

uradnky
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Homework Statement



Find the area bound by y= x - 2\sqrt{x} and y=0


The Attempt at a Solution



X intercept = 4
Y intercept = 0

Using vertical rectangles..

dA = ((Upper curve)-(Lower curve)) dx
\int(dA = \int(0-(x-2\sqrt{x})) dx

A= -(1/2X^2)-((2x^(3/2))/(2/3)) Evaluated from 0 to 4

A= 16

I understand this is an extremely easy question but I just cannot see where I have gone wrong with it. Is there a problem with my signs since the area I am finding is below the x-axis? Sorry if this is a bit messy, any help is appecricated, thanks.
 
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You should be calculating \int2sqrt(x) - x from 0 to 4.
 
That was easy. Thanks.
 

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