# How do you evaluate the definite integral by the limit definition given int (2x+5)dx from [0,2]?

Feb 25, 2017

Here is a limit definition of the definite integral. (I hope 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.)

Let's do one small step at a time.

${\int}_{0}^{2} \left(2 x + 5\right) \mathrm{dx}$.

Find $\Delta x$

For each $n$, we get

$\Delta x = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n}$

Find ${x}_{i}$

And ${x}_{i} = a + i \Delta x = 0 + i \frac{2}{n} = \frac{2 i}{n}$

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

$f \left({x}_{i}\right) = 2 \left({x}_{i}\right) + 5$

$= 2 \left(\frac{2 i}{n}\right) + 5$

$= \frac{4 i}{n} + 5$

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

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta t = {\sum}_{i = 1}^{n} \left(\frac{4 i}{n} + 5\right) \frac{2}{n}$

$= {\sum}_{i = 1}^{n} \left(\frac{8 i}{n} ^ 2 + \frac{10}{n}\right)$

$= \frac{8}{n} ^ 2 {\sum}_{i = 1}^{n} \left(i\right) + \frac{10}{n} {\sum}_{i = 1}^{n} 1$

Evaluate the sums

$= \frac{8}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right) + \frac{10}{n} \left(n\right)$

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

Rewrite before finding the limit

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = \frac{8}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right) + \frac{10}{n} \left(n\right)$

 = 4((n(n+1))/n^2)) + 10

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

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

To finish the calculation, we have

${\int}_{0}^{2} \left(2 x + 5\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} \left(4 \left(\frac{n \left(n + 1\right)}{n} ^ 2\right)\right) + 10$

$= 4 \left(1\right) + 10 = 14$