By Math Original No comments
Introduction to Integration

What is integration?

The process of integration is the reverse of the process of Differentiation. It is labeled by the symbol $ \displaystyle \int{,}$

Integration is used to find areas, volumes and it helps in a lot of other things. One of the most used is in finding the area of an unknown shape.

Example of an area problem:

In the figure below we have a region $ \displaystyle \Omega $ that is formed by a function $ \displaystyle f(x)$ and bounded by the lines $ \displaystyle x=a$ and $ \displaystyle x=b$.

The question is: What number should be called the area of $ \displaystyle \Omega $?

integration 1

To find out this question we start to split up the interval $ \displaystyle \left[ {a,b} \right]$ into a finite number of intervals.

$ \displaystyle \left[ {{{x}_{0}},{{x}_{1}}} \right],\left[ {{{x}_{1}},{{x}_{2}}} \right],……..\left[ {{{x}_{{n-1}}},{{x}_{n}}} \right]$ with $ \displaystyle a={{x}_{0}}<{{x}_{1}}<……..<{{x}_{n}}=b$

This breaks up the region $ \displaystyle \Omega $ into n subregions: $ \displaystyle {{\Omega }_{1}},{{\Omega }_{2}},……{{\Omega }_{n}}$

We can calculate the total area of $ \displaystyle \Omega $ by calculating the area of each subregions: $ \displaystyle {{\Omega }_{1}},{{\Omega }_{2}},……{{\Omega }_{n}}$ and adding up the result. The shape of each region is in rectangular form so its area is width x length. The problem is that the answer won’t be accurate.

But we can try and make the intervals a lot smaller in width  than the previous figure and see that the space that is left, the uncolored part is getting smaller.

integration

The area of all the sub-regions is getting near the real area of the region $ \displaystyle \Omega $. Now the answer is more accurately than before.

*And the smaller the width of the intervals, the more we approach to the real answer we are looking for. The only problem is that it may be a lot of intervals and we need to do a lot of adding up of their areas.

There is a simple way to find that area without a lot of work and the answer is obvious!

By finding its integral, that as we said above is the reverse of finding a derivative.

We know that the derivative of the function $ \displaystyle y=a{{x}^{n}}$ is $ \displaystyle {y}’=na{{x}^{{n-1}}}$.

Since we said that integration is reverse of derivation than it means that the integral of the function $ \displaystyle y=na{{x}^{{n-1}}}$ is $ \displaystyle \int{{na{{x}^{{n-1}}}=a{{x}^{n}}}}$.

That’s true but not super accurately!

Parts of an integral of a function

integration

The part of the integral symbol and the function needs no further explanation, the part that needs explanation is “dx”and“c”.

What is “dx” and “c”?

“dx” is the change of width of the intervals we created above. More scientifically is called the differential of the variable x.

“c” is the constant of integration.

Note! It’s important to never forget to put the constant of integration because a lot of different functions may have the same derivative but that does not mean that the integral would be the same.

Example 1: The derivative of $ \displaystyle y={{x}^{3}}+2$ and $ \displaystyle y={{x}^{3}}+10$ it´s the same $ \displaystyle {y}’=3{{x}^{2}}$.

But if we write the integration of this function $ \displaystyle \int{{3{{x}^{2}}={{x}^{3}}}}$ we clearly know that this is only one answer but we can have so much more other functions that have the same exact answer, that’s why we shouldn’t forget to put the constant of integration at the end.

Example 2: What is the$ \displaystyle \int{{\frac{1}{x}}}$?

From the derivative table we know that the derivative of $ \displaystyle \ln x$ is $ \displaystyle {\frac{1}{x}}$ so since integration is the inverse of derivation we get:

$ \displaystyle \int{{\frac{1}{x}}}=\ln x+c$

We have two types of integrals

Indefinite Integral

Indefinite integrals are defined without the upper or lower limits. It is labeled as  $ \displaystyle \int{{f(x)dx=F(x)+C}}$

Definite Integral

It’s an integral that contains the upper and lower limits. It is labeled as $ \displaystyle \int\limits_{a}^{b}{{f(x)dx}}$, where C is the constant of integration

Copyright   © Math Original