- #1
SheldonAustin
- 2
- 0
We are already introduced to finding the value of definite integral by the anti-derivative approach
\int_{a}^{b}f(x) dx = F(b) - F(a)
In this approach we find the anti-derivative F(x) of f(x) and then subtract F(a) from F(b) to get the value of the definite integral
Reimann Sums is yet another method to compute the definite integral value from first principles
Let f(x) be a function defined on the closed interval [a,b] then the definite interval of f(x) from a to b is given by
`int_a^bf(x)dx` = `lim_(n->oo)(b-a)/n` `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n` ( provided the limit exits )
Examples for solve riemann sums and integrals
`int_1^2(2x+5)dx`
Let f(x) = 2x+5 and [a,b] = [1,2]
Δx = `(b-a)/n` = `(2-1)/n` =`1/n`
f(x) = 2x+5
f(a+rΔ) = f(1 + r`(1)/n` ) =2 (1 + r`(1)/n` ) + 5
By the Formula
`int_a^bf(x)dx` = `lim_(n->oo)(b-a)/n` `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n` ( provided the limit exits )
`int_1^2(2x+5)dx` = `lim_(n->oo)(1)/n` `sum_(r=1)^n ``(2(1+ r/n)+5)`
= `lim_(n->oo)(1)/n` `sum_(r=1)^n (7+ 2r/n)` <br>
`=lim_(n->oo)(1)/n sum_(r=1)^n``(7) +lim_(n->oo)(1)/n (2/n)sum_(r=1)^n (r)`
=`lim_(x->oo)1/n(7n)+1/n(2/n)lim_(n->oo) (n)(n+1)/2'`
`='lim_(n->oo)(7) +lim_(n->oo) 1/n(2/n)(n)(n+1)/2'`
`= 7 + lim_(n->oo)(n+1)/n`
`= 7 + lim_(n->oo)(1+ 1/n)`
`= 7 + 1`
`= 8`
Ans is 8
Let us Validate the Answer by using the anti-derivative Approach
`int_1^2(2x+5)dx= (2(x/2)^2 +5x)1^2`
`='(2^2-1^2) +5(2-1)'`
`= (4-1) + 5`
`= 3 +5`
`= 8`
\int_{a}^{b}f(x) dx = F(b) - F(a)
In this approach we find the anti-derivative F(x) of f(x) and then subtract F(a) from F(b) to get the value of the definite integral
Reimann Sums is yet another method to compute the definite integral value from first principles
Let f(x) be a function defined on the closed interval [a,b] then the definite interval of f(x) from a to b is given by
`int_a^bf(x)dx` = `lim_(n->oo)(b-a)/n` `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n` ( provided the limit exits )
Examples for solve riemann sums and integrals
`int_1^2(2x+5)dx`
Let f(x) = 2x+5 and [a,b] = [1,2]
Δx = `(b-a)/n` = `(2-1)/n` =`1/n`
f(x) = 2x+5
f(a+rΔ) = f(1 + r`(1)/n` ) =2 (1 + r`(1)/n` ) + 5
By the Formula
`int_a^bf(x)dx` = `lim_(n->oo)(b-a)/n` `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n` ( provided the limit exits )
`int_1^2(2x+5)dx` = `lim_(n->oo)(1)/n` `sum_(r=1)^n ``(2(1+ r/n)+5)`
= `lim_(n->oo)(1)/n` `sum_(r=1)^n (7+ 2r/n)` <br>
`=lim_(n->oo)(1)/n sum_(r=1)^n``(7) +lim_(n->oo)(1)/n (2/n)sum_(r=1)^n (r)`
=`lim_(x->oo)1/n(7n)+1/n(2/n)lim_(n->oo) (n)(n+1)/2'`
`='lim_(n->oo)(7) +lim_(n->oo) 1/n(2/n)(n)(n+1)/2'`
`= 7 + lim_(n->oo)(n+1)/n`
`= 7 + lim_(n->oo)(1+ 1/n)`
`= 7 + 1`
`= 8`
Ans is 8
Let us Validate the Answer by using the anti-derivative Approach
`int_1^2(2x+5)dx= (2(x/2)^2 +5x)1^2`
`='(2^2-1^2) +5(2-1)'`
`= (4-1) + 5`
`= 3 +5`
`= 8`