# Use the definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit. f(x)= x^2 + sqrt(1+2x)     6≤ x ≤ 8

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You need to evaluate the area under the given curve, hence, you should find the limit of sum of areas of n rectangles as n approaches to infinite.

You need to consider the width of rectangle as `Delta x`  and the height of rectangle as `f(x_i) ` .

You need to evaluate in case of n rectangles such that:

`Delta x = (8-6)/n =gt Delta x = 2/n`

You need to evaluate the area of rectangle using the right points such that:

`A = f(x_i)*Delta x`

`x_i = 6 + (2i)/n`

Substituting `x_i`  in `f(x_i)`  yields:

`f(x_i) = (6 + (2i)/n)^2 + sqrt(1 + 2(6 + (2i)/n))`

`f(x_i) = (6 + (2i)/n)^2 + sqrt(1 + 12 + (4i)/n)`

`f(x_i) = (6 + (2i)/n)^2 + sqrt(13 + (4i)/n)`

`A = f(x_i)*Delta x`

`A = ((6 + (2i)/n)^2 + sqrt(13 + (4i)/n))*(2/n)`

Expanding the binomial and multiplying by `2/n`  yields:

`A =72/n + 24i/n^2 + 8i^2/n^3 + (2/n)*sqrt(13 + (4i)/n)`

Hence, evaluating the area under the curve yields:

`A = lim_(n-gtoo) sum_(i=1)^n (72/n + 24i/n^2 + 8i^2/n^3 + (2/n)*sqrt(13 + (4i)/n))`