Question

Area of the region bounded by the curves y=1-x^2, x+y+1=0, x-y-1=0

Answer 1

Answer:

7/3 ≈ 2.33333

Step-by-step explanation:

Drawing a diagram would help.

The line x+y+1=0 is the line through the points (0, -1) and (-1, 0).

The line x-y-1=0 is the line through the points (0, -1) and (1, 0).

The curve y=1-x² is a parabola through the points (-1, 0) and (1, 0), with vertex at (0, 1).

The region is then made up simply of two components: (i) the region between the parabola and the x-axis from x = -1 to x = 1; (ii) the triangle with vertices at (-1, 0), (1, 0) and (0, -1).

The area of the triangle region is easy as half base times height gives

  area of triangle region = 2 × 1 / 2 = 1.

For the area of the parabolic region, this is an integral from -1 to 1...

$\text{area of parabolic region}\\\\=\int_{-1}^1(1-x^2)\,dx\\\\=\bigl[x - \tfrac13x^3\bigr]_{-1}^1\\\\=(1-\tfrac13)-(-1+\tfrac13)\\\\=\frac43$

So in total, the required area is

  1 + 4/3 = 7/3

Hope this helps!