Setup the limit definition of definite integrals int_-1^2(1+x^2)dx ?
Thanks for your help!
Thanks for your help!
1 Answer
int_(-1)^2 \ (1+x^2) \ dx = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+((3i)/n-1)^2}
Which if we evaluate gives us
Explanation:
Using Riemann sums. By definition of an integral, then
int_a^b \ f(x) \ dx
represents the area under the curve
That is
int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)
And we partition the interval
Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) }
\ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b-1 }
Here we have
Delta = {-1, -1+1(3)/n, -1+2 (3)/n, -1+3 (3)/n, ..., 2 }
And so:
I = int_(-1)^2 \ (1+x^2) \ dx
\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(-1+(3i)/n)
\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+(-1+(3i)/n)^2}
\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+((3i)/n-1)^2}
Which is the desired limit.
We can continue and evaluate the limit (ie the integral) by using the standard summation formula:
sum_(r=1)^n r \ = 1/2n(n+1)
sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1)
So, we have:
I = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+(9i^2)/n^2-(6i)/n+1}
\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {2+(9i^2)/n^2-(6i)/n}
\ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n \ 2 + 9/n^2 sum_(i=1)^n \ i^2 - 6/n sum_(i=1)^n \ i }
\ \ = lim_(n rarr oo) 3/n {2n + 9/n^2 1/6n(n+1)(2n+1) - 6/n 1/2n(n+1) }
\ \ = lim_(n rarr oo) 3/n {1/(2n)(4n^2 + 3(n+1)(2n+1) - 6n(n+1) }
\ \ = lim_(n rarr oo) 3/n {1/(2n)(4n^2 + 3(2n^2+3n+1) - (6n^2+6n) }
\ \ = lim_(n rarr oo) 3/(2n^2) {4n^2 + 6n^2+9n+3 - 6n^2-6n }
\ \ = 3/2 \ lim_(n rarr oo) 1/(n^2) {4n^2 + 3n+3 }
\ \ = 3/2 \ lim_(n rarr oo) {4 + 3/n+3/n^2 }
\ \ = 3/2 \ {4 + 0+0 }
\ \ = 6
Using Calculus
If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:
I = int_(-1)^2 \ 1+x^2 \ dx
\ \ = [x+x^3/3]_(-1)^2
\ \ = (2+8/3)-(-1-1/3)
\ \ = 6