<p>This is because there’s a relationship between summation and integrals. An integral is basically a way how to calculate an area under the curve. With right sum you can do the same thing. For example, mathematically </p>

<p><span class="math-container">$$ \sum_{i=0}^n f(i) \approx \int_0^n f(x)dx $$</span></p>

<p>In fact under the Reimann’s definition of integral, integrals are literary defined as a limit of a sum of function output multiplied by change of x or:</p>

<p><span class="math-container">$$\int_a^b f(x)dx = \lim \sum_{i=0}^{n-1} f(\xi_i) \Delta x_i$$</span></p>

<p>Now to go back to your question you can imagine <span class="math-container">$p_i c_i$</span> as an area of rectangle of height <span class="math-container">$p_i$</span> and base <span class="math-container">$c_i$</span>. Consequently <span class="math-container">$\sum p_i c_i$</span> can be considered an adding of of all areas of these rectangles and <span class="math-container">$\int p_i c_i di $</span> is just continuous approximation of this area. </p>

CC BY-SA 4.0
2020-05-02T11:31:10.250Z
Many thanks. Just one more clarification to get integral from 0 to 1 you divide the area by n ?
Vishnu Kant
CC BY-SA 4.0
2020-05-02T11:37:00.003Z
@VishnuKant not the area. You get area from 0-1 by assuming there is continuum of goods that is normalized to 1. However, to normalize the continuum of goods to 1 you would divide by n (it’s like when you want to turn something into a percentage). But the continuum of goods is not area in itself - geometrically it would be just a line segment
1muflon1
CC BY-SA 4.0
2020-05-02T10:45:55.027Z
2020-05-02T10:45:55.027Z
1muflon1
<p>In our lecture, first we had expenditure of consumption basket as <span class="math-container">$\sum_{i=1}^{n} p_{i}c_{i} $</span> but later my professor converted it to <span class="math-container">$\int_{0}^{1}p_{i}c_{i}di$</span>. I want to understand how she did this? And how can I differentiate <span class="math-container">$\int_{0}^{1}p_{i}c_{i}di$</span> with respect <span class="math-container">$c_{i}$</span> </p>

<mathematical-economics>
converting sum of consumption basket to integral
Vishnu Kant
# converting sum of consumption basket to integral

(Asked by: username_0 on 2020-05-01)

In our lecture, first we had expenditure of consumption basket as $\sum\_{i=1}^{n} p\_{i}c\_{i} $ but later my professor converted it to $\int\_{0}^{1}p\_{i}c\_{i}di$. I want to understand how she did this? And how can I differentiate $\int\_{0}^{1}p\_{i}c\_{i}di$ with respect $c\_{i}$

**Tags:** mathematical-economics


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## 1 answers

### Answer by username_1 on 2020-05-02. Score: 1 | Accepted answer

This is because there’s a relationship between summation and integrals. An integral is basically a way how to calculate an area under the curve. With right sum you can do the same thing. For example, mathematically

$$ \sum\_{i=0}^n f(i) \approx \int\_0^n f(x)dx $$

In fact under the Reimann’s definition of integral, integrals are literary defined as a limit of a sum of function output multiplied by change of x or:

$$\int\_a^b f(x)dx = \lim \sum\_{i=0}^{n-1} f(\xi\_i) \Delta x\_i$$

Now to go back to your question you can imagine $p\_i c\_i$ as an area of rectangle of height $p\_i$ and base $c\_i$. Consequently $\sum p\_i c\_i$ can be considered an adding of of all areas of these rectangles and $\int p\_i c\_i di $ is just continuous approximation of this area.

> **Comments:**
> **username_0**: Many thanks. Just one more clarification to get integral from 0 to 1 you divide the area by n ?  
> **username_1**: @username_0 not the area. You get area from 0-1 by assuming there is continuum of goods that is normalized to 1. However, to normalize the continuum of goods to 1 you would divide by n (it’s like when you want to turn something into a percentage). But the continuum of goods is not area in itself - geometrically it would be just a line segment  

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