How do you evaluate the definite integral #int (x^2 + 1)dx# from # [1,2]#?

1 Answer
Mar 19, 2016

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:

#int_1^2 (x^2+1)dx#.

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 .

.#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#.

Where, for each positive integer #n#, we let #Deltax = (b-a)/n#

And for #i=1,2,3, . . . ,n#, we let #x_i = a+iDeltax#. (These #x_i# are the right endpoints of the subintervals.)

I prefer to do this type of problem one small step at a time.

#int_1^2 (x^2+1)dx#.

For each #n#, we get

#Deltax = (b-a)/n = (2-1)/n = 1/n#

And #x_i = a+iDeltax = 1+i1/n =1+ i/n#

#f(x_i) = (x_i)^2 +1= (1+i/n)^2+1#

# = (1+(2i)/n+i^2/n^2)+1 = i^2/n^2+(2i)/n+2#

#sum_(i=1)^n f(x_1)Deltax = sum_(i=1)^n(i^2/n^2+(2i)/n+2) 1/n#

# = 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,

#sum_(i=1)^n f(x_1)Deltax = 1/6 [(n(n+1)(2n+1))/n^3]+ 2/2[(n(n+1))/n^2]+2#

The last thing to do is evaluate the limit as #nrarroo#.
I hope it is clear that this amounts to evaluating
#lim_(nrarroo)(n(n+1)(2n+1))/n^3# and #lim_(nrarroo)[(n(n+1))/n^2]#

There are several ways to think about the first limit :

The numerator can be expanded to a plynomial with leading term #2n^3#, so the limit as #nrarroo# is #2#.

OR

#(n(n+1)(2n+1))/n^3 = (n/n)((n+1)/n)((2n+1)/n)#

The limit at infinity is #(1)(1)(2)=2# as a product of rational expressions.

OR

#(n(n+1)(2n+1))/n^3 = (n/n)((n+1)/n)((2n+1)/n)#

# = (1)(1+1/n)(2+1/n)#

So the limit is, again #2#.

The second limit

The ideas above can be applied to get

#lim_(nrarroo)[(n(n+1))/n^2] = 1#

Completing the integration

.#int_1^2(x^2+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#

# = 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 #x^2+1# (call it #F(x)#) and evaluate #F(2)-F(1)#

#int_1^2(x^2+1) dx = (x^3/3+x)]_1^2#

# = [(2)^3/3+2]-[(1)^3/3+(1)]#

# = 14/3-4/3=10/3#