Calculus Question Area Bounded

Question

What is the area of the region bounded by the graphs of y = x² and y = 1

Answer 1

First, we review the boundaries of the area. The lower bound is y=x^2 and the upper bound is y=1. That is the complement of the area under the parabola y=x^2.

There are different ways to look at it.

A more geometrical solution:

Because of it's symmetry we can look at the region for 0<=x<1.

The area under the parabola is the integral between 0 and 1 of x^2 = [1/3 x^3] evaluated at 1 and 0 = 1/3*1^3 - 1/3*0^3 = 1/3.

The area we are looking for is the complement of that, that is the total area of the square 1x1 = 1 - the said area. That gives you 1 - 1/3 = 2/3.

Because of symmetry, we need to add the other half of the area and we get 2 * 2/3 = 4/3.

Alternatively, we can solve for the integral of 1- x^2 between -1 and 1, yielding [x - 1/3*x^3] between -1 and 1, which gives us 1 - 1/3 - (-1 + 1/3) = 2 - 2/3 = 4/3.

Answer 2

See graph https://www.desmos.com/calculator/faowopwoy7

To find the area between the two graphs, x² and 1, we take the integral of 1-x² with respect to x bounded at some point.

First, we can find these bounds by seeing where y=1 intersects y=x². The x values at these intersections will be our bounds. From the graph above, we see they intersect at x = -1 and x = 1.

Now, we can take the integral of 1-x² with respect to x between 1 and -1.

∫_-1¹ 1-x² dx = x - (1/3) x³ Ι¹_-1 = [1 - (1/3)] - [-1 - (-1/3)] = 2/3 + 2/3 = 4/3

The answer is 4/3.

Why integrate 1 - x²?

If we integrate x², then we are effectively finding the area between x² and 0.

See https://www.desmos.com/calculator/ippb3einrp for the graph of this.

Since we need to find the inverse of this area with respect to the bounds, you may have noticed we can just subtract the area under x² from the area of the square. The total area contained in the bounds is 2, and the area under x² is 2/3. 2 - (2/3) is also 4/3. Hope this helps.

Answer 3

See the videa.