Question

Consider the function `f(x)= -(x-3)^2+4` how do you write an equation using a limit to determine the area enclosed by f(x) and the x-axis?

Answer 1

`A = lim_(nrarroo)sum_(i=1)^n (-(i4/n-2)^2+4) 4/n`

Explanation:

The curve intersects the `x` axis at `f(x) =0`

or `x=1` and `x=5`.

Cut the interval `[1,5]` into `n` pieces each of length `(5-1)/n = 4/n`

The right endpoints of the subintervals are `1+(4i)/n`.

We can approximate the area under the curve on the `i^(th)` interval using a rectangle of

base `4/n` and

height `f(1+(4i)/n) = -(1+(4i)/n-3)^2+4 = -(i4/n-2)^2+4`.

We then sum the areas of the rectangles `sum_(i=1)^n (-(i4/n-2)^2+4) 4/n`.

Finally, we take a limit as the subintervals get shorter and shorter (go to `0`). This is aso the limit as the number of rectangles increases without bound (`nrarroo`)

`A = lim_(nrarroo)sum_(i=1)^n (-(i4/n-2)^2+4) 4/n`