# How do you use the limit process to find the area of the region between the graph y=2x-x^3 and the x-axis over the interval [0,1]?

Nov 22, 2016

See below.

#### Explanation:

Here is a limit definition of the definite integral. (I'd guess it's the one you are using.)

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

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.

${\int}_{0}^{1} \left(2 x - {x}^{3}\right) \mathrm{dx}$.

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

$f \left({x}_{i}\right) = 2 \left({x}_{i}\right) - {\left({x}_{i}\right)}^{3} = \frac{2 i}{n} - {\left(\frac{i}{n}\right)}^{3} = 2 \frac{i}{n} - {i}^{3} / {n}^{3}$

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

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

$= {\sum}_{i = 1}^{n} \left(2 \frac{i}{n} ^ 2 - {i}^{3} / {n}^{4}\right)$

$= {\sum}_{i = 1}^{n} \left(2 \frac{i}{n} ^ 2\right) - {\sum}_{i = 1}^{n} \left({i}^{3} / {n}^{4}\right)$

$= \frac{2}{n} ^ 2 {\sum}_{i = 1}^{n} \left(i\right) - \frac{1}{n} ^ 4 {\sum}_{i = 1}^{n} \left({i}^{3}\right)$

Evaluate the sums

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

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

Rewrite before finding the limit

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

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

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

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

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

$\lim \left(n \rightarrow \infty\right) \left(\frac{{n}^{2} {\left(n + 1\right)}^{2}}{n} ^ 4\right) = 1$

To finish the calculation, we have

${\int}_{0}^{1} \left(2 x - {x}^{3}\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} \left(\left(\frac{n \left(n + 1\right)}{n} ^ 2\right) - \frac{1}{4} \left(\frac{{n}^{2} {\left(n + 1\right)}^{2}}{n} ^ 4\right)\right)$

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