# How do you use the limit process to find the area of the region between the graph y=16-x^2 and the x-axis over the interval [1,3]?

Dec 21, 2016

Here is a limit definition of the definite integral. (Others are possible.)

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$.

#### Explanation:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$.

Where, for each positive integer $n$, we let $\Delta x = \frac{b - a}{n}$

And for $i = 1 , 2 , 3 , . . . , n$, we let ${x}_{i} = a + i \Delta x$. (These ${x}_{i}$ are the right endpoints of the subintervals.)

I prefer to do this type of problem one small step at a time.

${\int}_{1}^{3} \left(16 - {x}^{2}\right) \mathrm{dx}$.

Find $\Delta x$

For each $n$, we get

$\Delta x = \frac{b - a}{n} = \frac{3 - 1}{n} = \frac{2}{n}$

Find ${x}_{i}$

And ${x}_{i} = a + i \Delta x = 1 + i \frac{2}{n} = 1 + \frac{2 i}{n}$

Find $f \left({x}_{i}\right)$

$f \left({x}_{i}\right) = 16 - {\left({x}_{i}\right)}^{2} = 16 - {\left(1 + \frac{2 i}{n}\right)}^{2}$

$= 16 - \left(1 + \frac{4 i}{n} + \frac{4 {i}^{2}}{n} ^ 2\right)$

$= 15 - \frac{4 i}{n} - \frac{4 {i}^{2}}{n} ^ 2$

Find and simplify ${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$ in order to evaluate the sums.

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = {\sum}_{i = 1}^{n} \left(15 - \frac{4 i}{n} - \frac{4 {i}^{2}}{n} ^ 2\right) \frac{2}{n}$

$= {\sum}_{i = 1}^{n} \left(\frac{30}{n} - \frac{8 i}{n} ^ 2 - \frac{8 {i}^{2}}{n} ^ 3\right)$

$= {\sum}_{i = 1}^{n} \left(\frac{30}{n}\right) - {\sum}_{i = 1}^{n} \left(\frac{8 i}{n} ^ 2\right) - {\sum}_{i = 1}^{n} \left(\frac{8 {i}^{2}}{n} ^ 3\right)$

$= \frac{30}{n} {\sum}_{i = 1}^{n} \left(1\right) - \frac{8}{n} ^ 2 {\sum}_{i = 1}^{n} \left(i\right) - \frac{8}{n} ^ 3 {\sum}_{i = 1}^{n} \left({i}^{2}\right)$

Evaluate the sums

$= \frac{30}{n} \left(n\right) - \frac{8}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right) - \frac{8}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right)$

(We used summation formulas for the sums in the previous step.)

Rewrite before finding the limit

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = \frac{30}{n} \left(n\right) - \frac{8}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right) - \frac{8}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right)$

$= 30 - 4 \left(\frac{n \left(n + 1\right)}{n} ^ 2\right) - \frac{4}{3} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right)$

Now we need to evaluate the limit as $n \rightarrow \infty$.

${\lim}_{n \rightarrow \infty} \left(\frac{n \left(n + 1\right)}{n} ^ 2\right) = 1$

${\lim}_{n \rightarrow \infty} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right) = 2$

To finish the calculation, we have

int_0^1 x^2 dx = lim_(nrarroo) (30 - 4((n(n+1))/n^2) - 4/3((n(n+1)(2n+1))/n^3)

$= 30 - 4 \left(1\right) - \frac{4}{3} \left(2\right)$

$= \frac{90}{3} - \frac{12}{3} - \frac{8}{3} = \frac{70}{3}$.