Please help me. Use the Definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

Question

f(x) = x^2+sqrt(1+2x), 6 <= x <= 8

Answer 1

Let us put it first into definite integral, then we convert it into limits. The definition of definite integral in terms of sum and limits is:

∫^b_a f(x) dx = lim_n→∞ Σⁿ_i=1 f(a+Δx•i) Δx

where Δx = (b-a)/n

f(x) = x²+(1+2x)^1/2

where 6 ≤ x ≤ 8.

That means:

a = 6

b = 8

In this case the f(x) in the interval 6 ≤ x ≤ 8 is above the x-axis, You can check it in any graphing utility. So you don't have to worry about negative area (or area below the x-axis)

Now we can have a definite integral of the area:

∫⁸₆ [x²+(1+2x)^1/2] dx

Then we convert it into limits:

Δx = (8-6)/n

Δx = 2/n

f(a+Δx•i) = (a+Δx•i )² + (1+ 2(a+Δx•i ))^1/2 = (6 + 2i /n)² +(1+2(6 + 2i /n))^1/2

= (6 + 2i/n)² + ( 1+ 12 + 4i/n)^1/2

= (6 + 2i/n)² + ( 13 + 4i/n)^1/2

Therefore:

∫⁸₆ x²+(1+2x)^1/2 dx = lim_n→∞ Σⁿ_i=1[(6+ 2i/n)² + ( 13 + 4i/n)^1/2]•(2/n)