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Which integral gives the area of the region in the first quadrant bounded by the x-axis, y = x2, and x + y = 2?

the integral from 0 to 2 of x squared minus the quantity 2 minus x, dx

the integral from 0 to 1 of x squared, dx plus the integral from 1 to 2 of the quantity 2 minus x, dx

the integral from 0 to 1 of x squared, dx minus the integral from 1 to 2 of the quantity 2 minus x, dx

the integral from 0 to 2 of the difference of 2 minus x and x squared, dx

Respuesta :

Answer: [tex]\displaystyle \int_{0}^{1}x^2dx + \int_{1}^{2}(2-x)dx[/tex]

Which is choice B

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Explanation:

See the diagram below.

x+y = 2 is the same as y = 2-x

f(x) = x^2 and g(x) = 2-x intersect at (1,1) which is where we split the integrals so we have two regions to worry about

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[tex]\displaystyle \int_{0}^{1}x^2dx[/tex] shown in red in the diagram represents the area under y = x^2 from x = 0 to x = 1

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[tex]\displaystyle \int_{1}^{2}(2-x)dx[/tex]shown in blue (same diagram) represents the area under y = 2-x from x = 1 to x = 2.

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Combining the two integrals gets us the total area bounded by

  • the x axis
  • y = x^2
  • x+y = 2
Ver imagen jimthompson5910

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