Area under a curve

[stopandthink]

[math]

f(x)=x^2

[/math]

How do I go about finding the area under x=[0,1,2,3,4]??

I'm a beginner at definite Integrals so please make an easy explanation. Thanks!

[elfmotat]

Are you being asked to actually integrate the function, or to approximate it with rectangles or trapezoids? I ask because you have a discrete set of points there, not a continuous domain.

[stopandthink]

[math] \int_{0}^{4}x^2 dx=? [/math] is this how you start it off?

No ones asking me, I'm simply trying to learn how to integrate on my own. I just chose those points for simplicity

Okay I think I figured it out..

So first i take the integral of my function [math]x^2[/math] which is [math]1/3x^3[/math]

then I plug in 4 for x and get 21.3

then I plug in 1 for x and get 0.3

then I subtracted 21.3 - .3 = 21

 

sweet!

 

[studiot]

You are nearly there. I have written out the process in detail.
First the general integral of x squared, without any specific values of x is called the indefinite integral.

 

 

The indefinite integral is a formula or algebraic expression

It allows us to do the algebraic manipulation and is usually what is given in tables of standard integrals.

 

Standard integrals are ones that are already worked out.

 

This indefinite integral always has an additional constant called the constant of integration.

It is usually called called C.

This constant may be any value but once set is the same throught that specific example of the integral.

 

 

So the indefinite integral of x squared in your example is

[math]\int {{x^2}} dx = \frac{{{x^3}}}{3} + C[/math]

 

The d'something' is always needed and should be the same as the variable.

So dx for a function of x, dt for a function of t and so on.

 

When we provide specific values for x we obtain what is known as the definite integral.

[math]\int\limits_{x = 4}^{x = 0} {{x^2}dx} [/math]

[math] = \int_0^4 {{x^2}} dx[/math]

Note the ways of writing this. The values of x are called the limits of integration.

 

 

Then we substitute the formula for our indefinite integral into the above and place this in square brackets.

We keep the limits of integration at the end of the square brackets.

 

[math] = \left[ {\frac{{{x^3}}}{3} + C} \right]_0^4[/math]

This is now an expression of algebra and we can substitute the values of x and we subtract the first value that we have placed at the bottom of the integral and brackets from the second value we have placed at the top.

[math] = \left( {\frac{{{4^3}}}{3} + C} \right) - \left( {\frac{{{0^3}}}{3} + C} \right)[/math]

[math] = \frac{{64}}{3} + C - 0 - C[/math]

[math] = \frac{{64}}{3}[/math]

The definite integral is always just a number

 

Note since the constant C is constant for any specific example it cancels out.

So we usually do not bother to write it down for definite integrals.

 

In this case the indefinite integral yields the area under the curve. Or more precisely the area between the curve and the x axis.

However this is not always the case.

[math]\int_{ - 4}^4 x dx = \left[ {\frac{{{x^2}}}{2}} \right]_{ - 4}^4[/math]

[math] = \frac{{{4^2}}}{2} - \frac{{{{\left( { - 4} \right)}^2}}}{2}[/math]

[math] = 0[/math]
This is not the area between the line and the x axis.

 

Areas above the x axis are considered positive and areas below the x axis considered negative.

 

 

So if the function crosses the x axis some parts of the intgral are negative and some are positive.

 

The definite integral adds up these 'areas' with regard to sign as shown above and so arrives at zero in the example.

[math] = \left[ {0 - \frac{{{{\left( { - 4} \right)}^2}}}{2}} \right] + \left[ {\frac{{{{\left( 4 \right)}^2}}}{2} - 0} \right][/math]

= - 8 + 8 = 0

[math] = 0[/math]

To obtain the whole or total area we split the integral into sections, between points where the function crosses the axis.

[math]\int_{ - 4}^4 x dx = \int_{ - 4}^0 x dx + \int_0^4 x dx[/math]
And then we take the modulus or absolute value of each area and add them up.

[math] = | - 8| + 8 = 16[/math]

Definite integral = (- 8) + 8 = 0

 

but

 

Area = | - 8| + 8 = 16

 

You should draw the curves above to help see what is meant.

 

 

Does this help?

Edited June 27, 2013 by studiot

[stopandthink]

Great! so now I understand how to find the definite integral of two positive points on the x axis. And also that I must add a constant to an Indefinite Integral. Although, I'll admit that I'm confused now with finding the definite integral of positive negative limits. Basically I just split it into parts, get the absolute value, and then add both parts, I think?

---------------------

hold on so if someone asked for the definite integral of your example would you give them 0 or 16 as the answer? I see that area is not the same as definite integral. Hmmm

[studiot]

so now I understand how to find the definite integral of two positive points on the x axis

 

Between two points, not of.

 

finding the definite integral of positive negative limits

 

It's not whether the limits are positive or negative,but whether the curve is above or below the x axis.

I'm sure I said that somewhere.

 

Did you draw the graph, it is an easy one?

 

I know what I wrote was a lot to take in, but I tried to make it comprehensive since you are learning by yourself - good on you.

[stopandthink]

Yes, it's just a simple straight line. Also, thank you for helping me out