Calculus I - Definition of the Definite Integral

Show All Notes Hide All Notes

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.

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