look at the graph of y = 3 + √(1-x^2) from -1 to 0
because of the symmetry we could go from 0 to 1.
so the area consists of a rectangle 3 units by 1 unit, with the curve of √(1-x^2) sitting on top
let's just find the area of that
= ∫ √(1-x^2) dx from 0 to 1
not sure if you know the procedure to integrate that but Wolfram can do it for us
http://www.wolframalpha.com/input/?i=integrate+%E2%88%9A(1-x%5E2)+dx+from+0+to+1
add the rectangle and you are done
Evaluate the integral by interpreting it in terms of areas.
0
(3 +sqrt(1 − x2)) dx
−1
2 answers
geometrically the argument makes sense. But algebraically,
∫[a,b] f(x) dx = -∫[b,a] f(x) dx
∫[a,b] f(x) dx = -∫[b,a] f(x) dx