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Show that using the right endpoints, the area bounded by $f(x)=e^{x}$ and the $x$ -axis, between $x=0$ and $x=1,$ is given by (a) $A=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} e^{k / n} .$ (b) Observe that the given sum is a geometric sum and it follows that $A=(e-1) \lim _{n \rightarrow \infty} \frac{e^{1 / n}}{n\left(e^{1 / n}-1\right)} \cdot(\text { c) If we let } x=1 / n$ show that this limit is equivalent to $A=(e-1) \lim _{x \rightarrow 0} \frac{x e^{x}}{e^{x}-1} \cdot($ d) Compute this limit by recalling that near $x=0, e^{x} \approx 1+x$.We defined a Riemann sum as $\sum_{i=1}^{n} f\left(c_{i}\right) \Delta x$ where we partitioned the interval $[a, b]$ into $n$ equal subintervals,that is, we used a regular partition. Note that as $n$, the number of subintervals, becomes infinite, $\Delta x,$ the width of each subinterval, approaches zero. So we could write $$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(c_{i}\right) \Delta x=\lim _{\Delta x \rightarrow 0} \sum_{i=1}^{n} f\left(c_{i}\right) \Delta x$$ (7)where $c_{i}$ could be any point in the $i^{\text {th }}$ subinterval. We could divide the interval into an irregular partition where the subintervals have different lengths. Let $\Delta x_{1}$ be the width of the first subinterval, $\Delta x_{2}$ be the width of the second subinterval, $\ldots,$ and $\Delta x_{n}$ be the width of the $n^{\text {th }}$ subinterval. The definition of a Riemann sum is now generalized to$$\sum_{i=1}^{n} f\left(c_{i}\right) \Delta x_{i}$$The next step is to take the limit. In order to exhaust the region under the curve, we require the width of each rectangle to approach 0. If there were some subintervals whose width did not approach zero, then there is no way the Riemann sum could represent the area of the region (why?). There is an easier way of saying this: let $\|\Delta\|$ be the length of the largest of all the subintervals, if it approaches $0,$ then the length of every subinterval will approach zero (why?), and we have$$\lim _{|\Delta| \rightarrow 0} \sum_{i=1}^{n} f\left(c_{i}\right) \Delta x$$ (8)as the generalization to (7).

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 5

Sigma Notation and Areas

Integrals

Missouri State University

Campbell University

Baylor University

Idaho State University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

07:25

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