# Use the form of the definition of the integral given in the theorem to evaluate the integral. b=4 a=0 (6-x^2)dx

The definition of integral is (limit definition) is,

`int_a^bf(x)dx = lim_(n->oo)sum_(n=1)^nf(x_1^*)Deltax`

our function is, `int_0^4(6-x^2)dx`

the range is 0 to 4,

therefore, `Deltax = (4-0)/n = 4/n`

therefore,  `x_i^* = (4i)/n`

`sum_(n=1)^nf(x_1^*)Deltax = sum_(n=1)^nf((4i)/n)*(4/n)`

`= sum_(n=1)^n(6 - ((4i)/n)^2)*(4/n)`

`= sum_(n=1)^n(6 - (16i^2)/n^2)*(4/n)`

`= sum_(n=1)^n24/n - (64i^2)/n^3`

`= sum_(n=1)^n24/n - sum_(n=1)^n(64i^2)/n^3`

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The definition of integral is (limit definition) is,

`int_a^bf(x)dx = lim_(n->oo)sum_(n=1)^nf(x_1^*)Deltax`

our function is, `int_0^4(6-x^2)dx`

the range is 0 to 4,

therefore, `Deltax = (4-0)/n = 4/n`

therefore,  `x_i^* = (4i)/n`

`sum_(n=1)^nf(x_1^*)Deltax = sum_(n=1)^nf((4i)/n)*(4/n)`

`= sum_(n=1)^n(6 - ((4i)/n)^2)*(4/n)`

`= sum_(n=1)^n(6 - (16i^2)/n^2)*(4/n)`

`= sum_(n=1)^n24/n - (64i^2)/n^3`

`= sum_(n=1)^n24/n - sum_(n=1)^n(64i^2)/n^3`

`= 24/nsum_(n=1)^n1 - 64/n^3sum_(n=1)^ni^2`

we know, `sum_(n=1)^ni^2 = (n(n+1)(2n+1))/6`

we get,

`=24/n * n - 64/n^3 * (n(n+1)(2n+1))/6`

`=24 - (32(n+1)(2n+1))/(3n^2)`

`=(72n^2 - 32(2n^2+3n+1))/(3n^2)`

`=(72n^2 - 64n^2-96n-32)/(3n^2)`

`=(8n^2-96n-32)/(3n^2)`

`int_0^4(6-x^2)dx = lim_(n->oo)sum_(n=1)^nf(x_1^*)Deltax`

`int_0^4(6-x^2)dx = lim_(n->oo)(8n^2-96n-32)/(3n^2) =8/3 `

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