How do you use limits to evaluate int(x^2+4x-2)dx from [1,4]?

Apr 26, 2017

Here is a limit definition of the definite integral. (I don't know if 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.

${\int}_{1}^{4} \left({x}^{2} + 4 x - 2\right) \mathrm{dx}$.

Find $\Delta x$

For each $n$, we get

$\Delta x = \frac{b - a}{n} = \frac{4 - 1}{n} = \frac{3}{n}$

Find ${x}_{i}$

And ${x}_{i} = a + i \Delta x = 1 + i \frac{3}{n} = 1 + \frac{3 i}{n}$

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

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

$= 1 + \frac{6 i}{n} + \frac{9 {i}^{2}}{n} ^ 2 + 4 + \frac{12 i}{n} - 2$

$= \frac{9 {i}^{2}}{n} ^ 2 + \frac{18 i}{n} + 3$

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

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = {\sum}_{i = 1}^{n} \left(\frac{9 {i}^{2}}{n} ^ 2 + \frac{18 i}{n} + 3\right) \frac{3}{n}$

$= {\sum}_{i = 1}^{n} \left(\frac{27 {i}^{2}}{n} ^ 3 + \frac{54 i}{n} ^ 2 + \frac{9}{n}\right)$

$= {\sum}_{i = 1}^{n} \frac{27 {i}^{2}}{n} ^ 3 + {\sum}_{i = 1}^{n} \frac{54 i}{n} ^ 2 + {\sum}_{i = 1}^{n} \frac{9}{n}$

$= \frac{27}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{2} + \frac{54}{n} ^ 2 {\sum}_{i = 1}^{n} i + \frac{9}{n} {\sum}_{i = 1}^{n} 1$

Evaluate the sums

$= \frac{27}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right) + \frac{45}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right) + \frac{15}{n} \left(n\right)$

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

Rewrite before finding the limit

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = \frac{27}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right) + \frac{54}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right) + \frac{9}{n} \left(n\right)$

$= \frac{27}{6} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right) + \frac{54}{2} \left(\frac{n \left(n + 1\right)}{n} ^ 2\right) + 9$

$= \frac{9}{2} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right) + 27 \left(\frac{n \left(n + 1\right)}{n} ^ 2\right) + 9$

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

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

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

To finish the calculation, we have

${\int}_{1}^{4} \left({x}^{2} + 4 x - 2\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} \left(\frac{9}{2} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right) + 27 \left(\frac{n \left(n + 1\right)}{n} ^ 2\right) + 9\right)$

$= \frac{9}{2} \left(2\right) + 27 \left(1\right) + 9$

$= 45$