Question

Summation to integral? calculus

Answer 1

In calculus, summation involves adding discrete values, while integration is akin to adding an infinite number of infinitesimally small products to find areas, volumes, etc. Integral calculus extends these concepts to continuous functions and is used, for example, to find position from acceleration by integrating twice.

Step-by-step explanation:

The transition from summation to integral in calculus is an important concept that bridges the discrete world with the continuous. When we summate, we're adding together values to find a total amount. In a similar vein, an integral can be thought of as a continuous sum, where we're adding up infinitely small products to calculate an area, volume, or other quantities.

When performing an integral, the dimensions of what we're integrating must be consistent. For example, if we're integrating velocity (v) with respect to time (t), we obtain a dimension that is the product of the two, such as distance. This principle is rooted in the requirement of dimensional consistency, analogously understood in the phrase "You can't add apples and oranges."

Understanding the relationship between discrete sums and continuous integrals is crucial for applications in various fields, such as physics where integrals can represent cumulative effects over time or space. For example, integrating an acceleration function gives us velocity, and further integration yields position. This process allows us to work backward from acceleration to position, utilizing the power of integral calculus.

Answer 2

`\displaystyle\sum_(k=1)^n\left(\frac{3k}n-1\right)^4\frac3n`

You should recognize the
`\frac3n=\frac{b-a}n` term on the right to represent the width of each equally-spaced subinterval of the integration interval
`[a,b]`. We don't know the endpoints' exact values just yet.


`n` represents the total number of subintervals. If we fix it to some manageably small number, we can get a better picture of what the
`\left(\frac{3k}n-1\right)^4` term represents.

If
`n=3`, the partial sum reduces to


`\displaystyle\sum_(k=1)^3\left(\frac{3k}3-1\right)^4\frac33=\sum_(k=1)^3(k-1)^4=0^4+1^4+2^4`

This suggests the sum represents the left-endpoint Riemann summation to find the integral of
`x^4`, and in particular, the integral seems to be taken over the interval [0, 3]. Taking the limit, we get


`\displaystyle\lim_(n\to\infty)\sum_(k=1)^n\left(\frac{3k}n-1\right)^4\frac3n=\int_0^3x^4\,\mathrm dx`