For the intergral of -(x^2)/3 +6 interval [0,3]
Rewrite it as a function of n using the right hand endpoint without any summation sign.
can anyone help me start this problem out?
To rewrite the given integral as a function of n using the right-hand endpoint without a summation sign, we need to understand that the integral represents the area under the curve of the function over the interval [0, 3]. We need to divide this interval into n equal subintervals of width Δx.
The width of each subinterval (Δx) can be calculated as (b - a) / n, where a and b are the limits of integration, in this case, 0 and 3, respectively. So, Δx = (3-0) / n.
Next, we need to find the right-hand endpoints for each subinterval. The first right-hand endpoint will be x1 = a + Δx = 0 + Δx. For the second subinterval, x2 = x1 + Δx, and so on, until the nth subinterval, xn = x(n-1) + Δx.
Now, let's rewrite the integral using these right-hand endpoints:
∫[-(x^2)/3 + 6]dx from 0 to 3 ≈ Δx [-(x1^2)/3 + 6 + -(x2^2)/3 + 6 + ... + -(xn^2)/3 + 6]
Substituting the values of xi (right-hand endpoints), the integral expression becomes:
∫[-(x^2)/3 + 6]dx from 0 to 3 ≈ Δx [-(0^2)/3 + 6 + -(Δx^2)/3 + 6 + ... + -((nΔx)^2)/3 + 6]
Now, we can substitute the value of Δx in terms of n:
∫[-(x^2)/3 + 6]dx from 0 to 3 ≈ [(3-0) / n] [-(0^2)/3 + 6 + -(((3/n)^2)/3) + 6 + ... + -(((n(3/n))^2)/3) + 6]
Simplifying further:
∫[-(x^2)/3 + 6]dx from 0 to 3 ≈ (3/n) [6 - ((3^2)/(3n^2)) - ((6^2)/(3n^2)) - ... - ((3^2)/(3n^2))]
Thus, the rewritten function of n, using the right-hand endpoint, without any summation sign, is:
f(n) = (3/n) [6 - (9/n^2) - (36/n^2) - ... - (9/n^2)]