00:01
In this question, we are asked to evaluate the following definite integral using the definition of the definite integral.
00:09
Recall that the definition of a definite integral is its limit as n goes to infinity of the sum and from, sorry, k from zero to n minus 1, f of ck times delta x.
00:32
Here delta x is the length of the interval divided by n in our case it's going to be 3 over n and k is the left point of each sub interval and here is basically here is the picture here is our interval 0 to 3 and we divide this interval into subintervals into n subintervals so the length of each subinterval is equal to delta x and is equal to 3 over n and the left end point of which sub interval is here and we're not going to take 3 because we are taking left end point of each sub interval right then 3 is not a left end point of any sub interval and the formula for k is k is equal to 3 i over n so where i is equal i is equal to 0 and so on on, sorry, k, it should be k here.
01:58
K is equal to zero, so on, and the last one is going to be n minus one.
02:02
When k is equal to zero, we are going to get zero, right? which is the left end point of the first sub -interval.
02:08
When k equals one, we are going to get 3 over n, which is the left end point of the second sub -interval, and so on.
02:16
Therefore, and f is our function.
02:19
Here is f of x, 4x plus 5.
02:23
Therefore, we can rewrite our integral as limit as n goes to infinity of the sum k from 0 to n minus 1.
02:36
Now, f is equal to 4x plus 5, and we are going to replace x by k, which is 3k over n, plus 5, multiplied by delta x by delta x is equal to 3 over n.
02:55
Therefore, the problem of calculating this integral reduces to the problem of calculating this limit.
03:03
Let's simplify this a little bit.
03:06
You're going to get limit sum of 12k over n plus 5 times 3 over n.
03:20
Now we can simplify it even more.
03:23
You can rewrite this as limit.
03:27
N goes to infinity, sum of 36k over n squared plus 15 over n...