Math  /  Calculus

QuestionUse this definition with right endpoints to find an expression for the area under the graph of ff as a limit. Do not evaluate the limit. f(x)=xx3+5,1x4A=limni=1n\begin{array}{l} f(x)=x \sqrt{x^{3}+5}, 1 \leq x \leq 4 \\ A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \square \end{array} Need Help? Read It

Studdy Solution

STEP 1

1. We are given a function f(x)=xx3+5 f(x) = x \sqrt{x^3 + 5} and an interval [1,4][1, 4].
2. We need to express the area under the curve as a limit of a Riemann sum using right endpoints.
3. We will not evaluate the limit, only express it.

STEP 2

1. Divide the interval [1,4][1, 4] into nn subintervals.
2. Determine the width of each subinterval.
3. Identify the right endpoint of each subinterval.
4. Set up the Riemann sum using the right endpoints.
5. Express the area as a limit of the Riemann sum.

STEP 3

Divide the interval [1,4][1, 4] into nn subintervals. The width of each subinterval, Δx\Delta x, is given by:
Δx=41n=3n \Delta x = \frac{4 - 1}{n} = \frac{3}{n}

STEP 4

Determine the right endpoint of each subinterval. For the ii-th subinterval, the right endpoint xix_i is:
xi=1+iΔx=1+i(3n) x_i = 1 + i \Delta x = 1 + i \left(\frac{3}{n}\right)

STEP 5

Set up the Riemann sum using the right endpoints. The Riemann sum SnS_n is:
Sn=i=1nf(xi)Δx=i=1n((1+i3n)(1+i3n)3+5)3n S_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left( \left(1 + i \frac{3}{n}\right) \sqrt{\left(1 + i \frac{3}{n}\right)^3 + 5} \right) \frac{3}{n}

STEP 6

Express the area AA as a limit of the Riemann sum:
A=limni=1n((1+i3n)(1+i3n)3+5)3n A = \lim_{n \to \infty} \sum_{i=1}^{n} \left( \left(1 + i \frac{3}{n}\right) \sqrt{\left(1 + i \frac{3}{n}\right)^3 + 5} \right) \frac{3}{n}

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