How do you evaluate the definite integral #int (x^2 + 1)dx# from # [1,2]#?
1 Answer
For evaluation using the definition, please see the full explanation section below. (Skip to the end if you can use the Fundamental Theorem of Calculus.)
Explanation:
I am assuming that you do not yet have the Fundamental Theorem of Calculus available to evaluate this, but that you need to evaluate it from a definition .
.
Where, for each positive integer
And for
I prefer to do this type of problem one small step at a time.
For each
And
# = (1+(2i)/n+i^2/n^2)+1 = i^2/n^2+(2i)/n+2#
# = sum_(i=1)^n(i^2/n^3+(2i)/n^2+2/n)#
# = 1/n^3 sum_(i=1)^n i^2 + 2/n^2 sum_(i=1)^n i+ 2/n sum_(i=1)^n 1 #
# = 1/n^3[(n(n+1)(2n+1))/6] + 2/n^2[(n(n+1))/2]+2/n[n]#
(We used summation formulas for the three sums in the previous step.)
So,
The last thing to do is evaluate the limit as
I hope it is clear that this amounts to evaluating
There are several ways to think about the first limit :
The numerator can be expanded to a plynomial with leading term
OR
The limit at infinity is
OR
# = (1)(1+1/n)(2+1/n)#
So the limit is, again
The second limit
The ideas above can be applied to get
Completing the integration
.
# = lim_(nrarroo) sum_(i=1)^n(i^2/n^2+(2i)/n+2) 1/n#
# = lim_(nrarroo) [1/6[(n(n+1)(2n+1))/n^3]+[(n(n+1))/n^2]+[2]]#
# = 1/6 [2] + [1] + [2]#
# = 1/3+3 = 10/3#
Using the Fundamental Theorem of Calculus
Find an antiderivative of
# = [(2)^3/3+2]-[(1)^3/3+(1)]#
# = 14/3-4/3=10/3#