Indefinite Integral Explain?

Question

I don't understand it at all. In this picture, I have the two first rules:

http://s1290.photobucket.com/user/photoman164/media/math_zpse5d8cb7d.jpg.html?filters[user]=136734335&filters[recent]=1&sort=1&o=0

Can anyone explain these rules? I know it must sound simple, but I don't understand.

For the first rule, d just disappears and replaced with x. Why? I know that all indefinite integral problems just add a plus c at the end of the solution.

For the second rule, not sure why + 1 is added to n and why it also appears in the denominator as well. Any help appreciated. Thanks.

Answer 1

•       First rule: ∫ k dx = k⋅x + C



You know that ∫ just means ∑, except instead of a finite number of terms there is an infinite number of them. If you expand the sum out like this:



        ∫ k dx = k dx + k dx + k dx + k dx + ... = k⋅( dx + dx + dx + ... )



You can easily see that we can extract the 'k' part out -- it's in every term. So:



        ∫ k dx = k ∫ dx



That much should be clear.



Now all you need to know is that dx is an infinitesimal part of some finite number x. If you divide up x into an infinite number of infinitesimal parts, each called dx, then if you add back up these infinitesimal parts you of course get back x again. It's that simple. So, ∫ dx = x (except for a constant factor.)



A way to understand why indefinite integrals result in a +C part is to think of things a little differently. If you were driving a car down a road and used a computer to continually log the speedometer reading, you could use that series of continuous readings to figure out how far you drove. You'd just add up all the individual bits of speed⨯time to get the total distance driven. Problem is, that wouldn't tell you where you wound up, unless you also knew where you started from. So the "differentials" can be added up to figure out for far you went, but they cannot tell you where you ended up at without one more piece of info. This is the +C bit and so ∫ dx = x + C.



        ∫ k dx = k ∫ dx = k⋅(x + C₀) = k⋅x + k⋅C₀ = k⋅x + C₁



(Where C₁ = k⋅C₀.)



Think of the little 'd' just before 'x' as a special marker. In regular algebra, x and y and z are variables that may hold any finite value. In fact, all of the variables you use in algebra can ONLY hold finite values of some kind.



In calculus, you create a NEW kind of variable that can ONLY hold infinitely small, but non-zero, values. Instead of holding 1 or 2 or 5, they instead hold 1 ⁄ ∞ or 2 ⁄ ∞ or 5 ⁄ ∞. Like that. And so that you can tell which kind of variable is being discussed, you either do not (or do) place a small 'd' in front. So x can only hold finite values and dx can only hold infinitesimal values. But they are both just variables and you can cancel them out just like in regular algebra. You can divide both sides of an equation by 'dt' or multiply both sides by 'dy'. Just like you may in algebra with regular variables. It's just another kind of variable.



So when you sum up all the infinitesimal dx values, you always get back some finite x value. Because that finite x value was the value that was divided up into infinitely small pieces in the first place. It must be the case that if you divide up x into infinitely small dx pieces, the sum of all those dx pieces gets you back to x.



It's really nifty once you get the idea. Suddenly calculus is trivial and only slightly more complex than algebra was. You get to use all of what you learned in algebra with these weird infinitesimal variables, too. You can divide one by another to get a ratio that is finite, too. So if you see dy ⁄ dx = 5, you know that it must also be that dy = 5⋅dx or that dy is 5 times larger than dx is. They are both infinitely small -- true -- but that doesn't mean you can't figure out their ratio just like in algebra.



You can also tell immediately that it is "illegal" to write: x = 4⋅dy. Why? Because x is finite and dy is infinitesimal. And it is impossible to multiply an infinitesimally small value by 4 and get a finite value. It just cannot work.



When you read:



        ∫ x dx



You should "see" that x is being multiplied by dx to form an area that is x high and dx wide. The ∫ part is just summing up all those areas. Now, the areas must be infinitesimal because when you multiply a finite x by an infinitesimal value dx, the result itself must also be infinitesimally small. Which is why you need an ∫ type of sum to sum them up to make a finite value.



What does this sum "look like?" Well, let's say you want ∫₀⁵ x dx. Then your first "area" is at x=0 and is dx wide. So it is 0 high and dx wide. (Small). The next one will be at x=dx. So the 2nd area is dx high and dx wide. The third area is 2dx high and dx wide. Etc. All the way out until you reach the very last area which will be 5 high and dx wide. Put these areas side by side, okay? You get a triangle, right? It starts at x=0 and gets successively higher and higher until you reach x=5 where the last area is 5 high. Obviously, this triangle has a base of 5 and a height of 5 and will have an area of ²⁵⁄₂.



In calculus, ∫₀⁵ x dx = ½ x² |₀⁵ = ²⁵⁄₂



Same answer. But you can think geometrically, too. You don't have to only think algebraically or the way they like to teach you. You can learn to "see."



I'll stop there. You need to think about this a bit, I suspect.

Answer 2

Correct me if I've misread you, but what you seem not to understand is what the notation is; what an integral is?



When you integrate a function (taking the indefinite integral), you are "anti-differentiating" it; you're finding the function whose derivative you've placed inside the integral.

Another way to picture that is, given a graph of

y = f(x)

you're finding the area "under" the graph, from some fixed starting x-value, x=a, to a general x-value, and expressing the result as a function of that final x-value.



Rule#1 says that if you do that with y=f(x)=k (a constant), you're looking at the area of a rectangle whose width goes from from a to x, and whose height is k, and that you will get kx + some constant.



The notation is meant to preserve the essence of the limiting process that defines the integral -- you take

n

∑ f(x_k) ∆x

k=1



in which you run the x's from x=a to x=b,

x_0 = a

x_j = x_0 + j ∆x

x_n = b,



subdividing that interval into n little ∆x's,

∆x = (b - a)/n

and then you take the limit of that as n→∞.



That limit gets put into symbols as:



b

∫ f(x) dx

a



and the "dx" is the "infinitesimal" ∆x that went in the limit to 0.

So the "d" isn't separate; "dx" is a single symbol in this notation.



Rule#2, which husoski correctly points out, includes Rule#1 as a special case, with n=0, can be derived as he has done, using the Fundamental Theorem of Calculus



∫ f'(x) dx = f(x) + C



where C is an arbitrary constant, which becomes pinned down to a definite value when you choose definite limits of integration; i.e., take a definite integral:

b

∫ f'(x) dx = f(b) - f(a)

a



EDIT:

Caveat: Rule#2 is good for all real values of n, except n = -1 (where the formula would have you dividing by 0).

For that, you get a logarithmic function.



∫ xˉ¹ dx = ∫ (1/x) dx = ln |x|

Answer 3

First off, the constant rule is just the power rule used with a constant times x^0, so there's really only one rule here.



Second, the power rule for integration comes from applying the Fundamental Theorem to the power rule for derivatives. Remember that:



d/dx [ x^n ] = n x^(n-1)



On form of the Fundamental Theorem states that:



∫ f'(x) dx = f(x) + C



If you can find some f(x) such that f'(x) = x^n, then you have your indefinte integral.



Change that derivative power rule a bit:



d/dx [ x^(n+1) ] = (n+1) x ^n



That just use one more in the exponent, but the rule still works. Divide both sides by (n+1) and move the division by a constant inside the derivative on the left:



d/dx [ x^(n+1) / (n + 1) ] = x^n



There you are. If f(x) = x^(n+1) / (n + 1) then f'(x) = x^n. Then the F.T. gives:



∫ x^n dx = x^(n + 1) / (n + 1) + C



Since the derivative rule works for any value of n, integer or not, the integral rule works for all n with the lone exception of n=-1, where zero division occurs. In an odd twist, the integral of (1/x) dx will turn out to be the natural logarithm, but that's another story.