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

Question

Christopher Allers · Accepted 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.)

#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#

Scarlett Allison · Answer

undefined To evaluate the definite integral $\int_{1}^{2} (x^2 + 1)dx$, we first find the antiderivative of the integrand, which is $\frac{x^3}{3} + x$.

Next, we substitute the upper limit, 2, into the antiderivative and subtract the result of substituting the lower limit, 1:

$$\left(\frac{2^3}{3} + 2\right) - \left(\frac{1^3}{3} + 1\right)$$

$$= \left(\frac{8}{3} + 2\right) - \left(\frac{1}{3} + 1\right)$$

$$= \left(\frac{8}{3} + \frac{6}{3}\right) - \left(\frac{1}{3} + \frac{3}{3}\right)$$

$$= \frac{14}{3} - \frac{4}{3}$$

$$= \frac{10}{3}$$

So, $\int_{1}^{2} (x^2 + 1)dx = \frac{10}{3}$.