# How do you use the limit process to find the area of the region between the graph y=x^2+2 and the x-axis over the interval [0,1]?

Nov 13, 2016

Please see the explanation section below.

#### Explanation:

Here is a limit definition of the definite integral. (I'd guess it's the one you are using.) I will use what I think is somewhat standard notation in US textbooks.

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

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.

${\int}_{0}^{1} \left({x}^{2} + 2\right) \mathrm{dx}$.

Find $\Delta x$

For each $n$, we get

$\Delta x = \frac{b - a}{n} = \frac{1 - 0}{n} = \frac{1}{n}$

Find ${x}_{i}$

And ${x}_{i} = a + i \Delta x = 0 + i \frac{1}{n} = \frac{i}{n}$

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

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

$= {i}^{2} / {n}^{2} + 2$

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

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

$= {\sum}_{i = 1}^{n} \left({i}^{2} / {n}^{3} + \frac{2}{n}\right)$

$= {\sum}_{i = 1}^{n} \left({i}^{2}\right) + {\sum}_{i = 1}^{n} \left(\frac{2}{n}\right)$

$= \frac{1}{n} ^ 3 {\sum}_{i = 1}^{n} \left({i}^{2} / {n}^{3}\right) + \frac{2}{n} {\sum}_{i = 1}^{n} \left(1\right)$

Evaluate the sums

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

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

Rewrite before finding the limit

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

$= \frac{1}{6} \left(\frac{n}{n} \cdot \frac{n + 1}{n} \cdot \frac{2 n + 1}{n}\right) + 2$

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

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

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

${\lim}_{n \rightarrow \infty} \frac{2 n + 1}{n} = 2$

To finish the calculation, we have

${\int}_{0}^{1} \left({x}^{2} + 2\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} \left(\frac{1}{6} \left(\frac{n}{n} \cdot \frac{n + 1}{n} \cdot \frac{2 n + 1}{n}\right) + 2\right)$

$= \frac{1}{6} \left(\left(1\right) \left(1\right) \left(2\right)\right) + 2$

$= \frac{1}{3} + 2 = \frac{7}{3}$.