# Use the Definition to find an expression for the area under the graph of fas a limit. Do not evaluate the limit. Use the Definition to find an expression for the area under the graph of fas a limit. Do not evaluate the limit. f(x) = x2+ sqrt(1+2x) , 4 ≤ x ≤ 6 lim n---> infinity Sigma, with i = 1 and n on top

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You need to evaluate the definite integral of the given funtion to find the area under its graph such that:

`int_1^n (x^2+ sqrt(1+2x)) dx `

You need to use the property of linearity of integrals such that:

`int_1^n (x^2+ sqrt(1+2x)) dx = int_1^n (x^2)dx+ int_1^n sqrt(1+2x)) dx `

`int_1^n (x^2+ sqrt(1+2x)) dx = x^3/3|_1^n + int_1^n sqrt(1+2x)) dx `

You need to use the following substitution to solve `int_1^n sqrt(1+2x)) dx`  such that:

`2x+1 = t => 2dx = dt`

`int sqrt(1+2x)dx = int sqrt t dt/2 `

Converting the square root into a power yields:

`int sqrt t dt/2 = int t^(1/2) dt/2 = (1/2)(t^(1/2+1))/(1/2+1) + c`

`int sqrt t dt/2 = (t^(3/2))/3 + c => int_1^n sqrt(1+2x)dx = (1+2x)^(3/2)/3|_1^n`

`int_1^n (x^2+ sqrt(1+2x)) dx = (n^3 + (1+2n)^(3/2) - 1^3 - (1+2)^(3/2))/3`

`int_1^n (x^2+ sqrt(1+2x)) dx = (n^3 + (1+2n)^(3/2) - 1 - 3sqrt3)/3`

Hence, evaluating the area under the graph, between the limits 1 and n, yields `int_1^n (x^2+ sqrt(1+2x)) dx = (n^3 + (1+2n)^(3/2) - 1 - 3sqrt3)/3.`

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