Find the area bounded by the curves, y=x^2, y=3-x, y=0

Answer 1

y=3-x

y=x^2

x^2 = 3-x

x^2+x = 3

x^2 +x + 1/4 = 3+ 1/4

(x+1/2)^2 = 13/4

x + 1/2 = + (1/2)sqr13

x=(sqr13-1)/2 = 1.3 (ignore the negative square root, it's where the parabola intersects the line in quadrant II)

Area = integral of x^2 evaluated from 0 to (sqr13-1)/2 = 0.74

plus area of a right triangle = (1/2)bh = .5(1.3)(1.7) = 1.11

1.11+ 0.74 = 1.85

There may be an ambiguity in the problem. y=0 seems to require that the x-axis is one side of the area, so that the area is entirely in quadrant I. with the upper boundary the parabola only between x=0 and x= about 1.3 and the upper boundary the line 3-x between x=1.3 and x=3.

However, if you ignored y=0 and looked only for the area between the parabola and the line, you'd get a different area, mostly in quadrant II. That area could have just as easily been described as bounded by y=any negative number, making y=0 superfluous or irrelevant to the problem. Because y=0 is an explicit boundary. the required area seems to be the area in quadrant I, with area = about 1.85

Answer 2

First find the intersection point (which will be the limits of integration.

If y = x² and y = 3 - x the x² = 3 - x

x² + x - 3 = 0

x = (-1 ± √[1² - 4(1)(-3)])/[2(1)] = (-1 ± √13)/2

So the limits of integration are -1/2 - √13/2 and -1/2 + √13/2

So the area between the curves is ∫(3-x)dx - ∫x² dx evaluated between these limits of integration.

First Part:

∫(3-x)dx = 3x - x²/2 and evaluated between -1/2 - √13/2 and -1/2 + √13/2 gives:

3(-1/2 + √13/2) - (-1/2 + √13/2)²/2 - [3(-1/2 - √13/2) - (-1/2 - √13/2)²/2]

-3/2 + 3√13/2 - (1/4 - √13/2 + 13/4)/2 - [-3/2 - 3√13/2 - (1/4 + √13/2 + 13/4)/2)]

-12/8 + 6√13/4 - 1/8 + √13/4 - 13/8 - [-12/8 - 6√13/4 - 1/8 - √13/4 - 13/8]

-26/8 +7√13/4 - (-26/8 - 7√13/4)

-26/8 +7√13/4 + 26/8 + 7√13/4

14√13/4 = 7√13/2

Second Part:

∫x² dx = x³/3 and evaluated between -1/2 - √13/2 and -1/2 + √13/2 gives:

(-1/2 + √13/2)³/3 - (-1/2 - √13/2)³/3

(-5 + 2√13)/3 - (-5 - 2√13)/3

-5/3 + 2√13/3 - (-5/3 - 2√13/3)

-5/3 + 2√13/3 + 5/3 + 2√13/3)

4√13/3

First Part - Second Part = 7√13/2 - 4√13/3 = 21√13/6 - 8√13/6 = 13√13/6

So the area between the curves is 13√13/6 (which is approx 7.812)

Answer 3

The area can be found by integrating between the points of intersection for y=3-x and y=x² using for the integrand the difference (upper function)-(lower function). The video goes through all the steps for finding the solution.

Best of Luck.

David