Paul's Online Notes
Home / Calculus I / Integrals / Definition of the Definite Integral
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 5.6 : Definition of the Definite Integral

1. Use the definition of the definite integral to evaluate the integral. Use the right end point of each interval for $$x_{\,i}^*$$.

$\int_{1}^{4}{{2x + 3\,dx}}$

Show All Steps Hide All Steps

Start Solution

The width of each subinterval will be,

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

The subintervals for the interval $$\left[ {1,4} \right]$$ are then,

$\left[ {1,1 + \frac{3}{n}} \right],\left[ {1 + \frac{3}{n},1 + \frac{6}{n}} \right],\left[ {1 + \frac{6}{n},1 + \frac{9}{n}} \right], \ldots ,,\left[ {1 + \frac{{3\left( {i - 1} \right)}}{n},1 + \frac{{3i}}{n}} \right], \ldots ,,\left[ {1 + \frac{{3\left( {n - 1} \right)}}{n},4} \right]$

From this it looks like the right end point, and hence $$x_i^*$$, of the general subinterval is,

$x_i^* = 1 + \frac{{3i}}{n}$ Show Step 2

The summation in the definition of the definite integral is then,

$\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \sum\limits_{i = 1}^n {\left[ {2\left( {1 + \frac{{3i}}{n}} \right) + 3} \right]\left[ {\frac{3}{n}} \right]} = \sum\limits_{i = 1}^n {\left[ {\frac{{15}}{n} + \frac{{18i}}{{{n^2}}}} \right]}$ Show Step 3

Now we need to use the formulas from the Summation Notation section in the Extras chapter to “evaluate” the summation.

\begin{align*}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} & = \sum\limits_{i = 1}^n {\frac{{15}}{n}} + \sum\limits_{i = 1}^n {\frac{{18i}}{{{n^2}}}} = \frac{1}{n}\sum\limits_{i = 1}^n {15} + \frac{{18}}{{{n^2}}}\sum\limits_{i = 1}^n i \\ & = \frac{1}{n}\left( {15n} \right) + \frac{{18}}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) = 15 + \frac{{9n + 9}}{n}\end{align*} Show Step 4

Finally, we can use the definition of the definite integral to determine the value of the integral.

$\int_{1}^{4}{{2x + 3\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \mathop {\lim }\limits_{n \to \infty } \left[ {15 + \frac{{9n + 9}}{n}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {24 + \frac{9}{n}} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{24}}$