# 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?

Aug 14, 2017

$A = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left(- {\left(i \frac{4}{n} - 2\right)}^{2} + 4\right) \frac{4}{n}$

#### Explanation:

The curve intersects the $x$ axis at $f \left(x\right) = 0$

or $x = 1$ and $x = 5$.

Cut the interval $\left[1 , 5\right]$ into $n$ pieces each of length $\frac{5 - 1}{n} = \frac{4}{n}$

The right endpoints of the subintervals are $1 + \frac{4 i}{n}$.

We can approximate the area under the curve on the ${i}^{t h}$ interval using a rectangle of

base $\frac{4}{n}$ and

height $f \left(1 + \frac{4 i}{n}\right) = - {\left(1 + \frac{4 i}{n} - 3\right)}^{2} + 4 = - {\left(i \frac{4}{n} - 2\right)}^{2} + 4$.

We then sum the areas of the rectangles ${\sum}_{i = 1}^{n} \left(- {\left(i \frac{4}{n} - 2\right)}^{2} + 4\right) \frac{4}{n}$.

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 ($n \rightarrow \infty$)

$A = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left(- {\left(i \frac{4}{n} - 2\right)}^{2} + 4\right) \frac{4}{n}$