# Set up ʃ (superscript 6)(subscript 2) e^x sin x dx as the limit of a Riemann Sum.Set up but do not evaluate ʃ (superscript 6)(subscript 2) e^x sin x dx as the limit of a Riemann Sum. You can...

Set up ʃ (superscript 6)(subscript 2) e^x sin x dx as the limit of a Riemann Sum.

Set up but do not evaluate ʃ (superscript 6)(subscript 2) e^x sin x dx as the limit of a Riemann Sum. You can choose x(superscript *)(subscript i) as right endpoints of the interaval [x(subscript i),x(subscript i+1)].

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### 2 Answers

You need to divide the width of interval `[2,6]` into n parts such that:

`(6-2)/n = 4/n`

You should select x_k as the right point of interval such that:

`f(x_k) = e^(x_k)sin(x_k)`

Substituting `4k/n` for `x_k` yields:

`f(4k/n) = e^(4k/n)sin(4k/n)`

You need to evaluate the Riemann sum such that:

`S_n = sum_(k=1)^n f(x_k)*(4/n)`

`S_n = sum_(k=1)^n e^(4k/n)sin(4k/n)*(4/n)`

**Hence, setting `int_2^6 e^xsin x dx` as a Riemann sum yields `S_n = sum_(k=1)^n e^(4k/n)sin(4k/n)*(4/n). ` **

**Sources:**

Evaluate `int_2^6e^xsinxdx` as a limit of Riemann sums:

Using a regular partition we get `Delta_(x_i)=(6-2)/n=4/n`

(A regular partition creates subintervals of equal width)

Using right endpoints of each interval yields `x_i^"*"=2+(4i)/n`

(Start at 2: each iteration move 4/n units)

The Riemann sum is `sum_(i=1)^nf(x_i^"*")Deltax_i^"*"` so we get:

`sum_(i=1)^n f(2+(4i)/n)4/n=sum_(i=1)^n (e^((2+(4i)/n))sin(2+(4i)/n))4/n`

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**The area indicated by `int_2^6e^xsinxdx` can be found by:**

**`lim_(n->oo)sum_(i=1)^n (e^((2+(4i)/n))sin(2+(4i)/n))4/n` **

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**Sources:**