How to Calculate Definite Integrals?

[clm222]

I made a pdf so that the equation would be more clear

 

[Muphrid]

Calculating the antiderivative and evaluating it on the bounds is almost always how definite integrals are done. The fundamental theorem of calculus gives an exact result in this way. Are you sure you didn't make some mistake either in finding the antiderivative or the Riemann sum?

Also, did you use LaTeX for that? If so, you can use inline and display math between itex or tex tags. Quote this post to see an example:

\int_a^b f(x) \; dx = F(b) - F(a)

Edit: though I should say there's an entire field concerning numerical integration--using Riemann sums or improvements on those techniques--to deal with functions whose antidervatives can't be found analytically.

 

[HallsofIvy]

The only purposes of Riemann sums are to give a basic definition of the Riemann integral and, sometimes, to reason out how to set up an integral. You never use Riemann sums to actually do an integral. Find an anti-derivative, evaluate it at the upper and lower bounds, and subtract.

 

[arildno]

The only purposes of Riemann sums are to give a basic definition of the Riemann integral and, sometimes, to reason out how to set up an integral. You never use Riemann sums to actually do an integral. Find an anti-derivative, evaluate it at the upper and lower bounds, and subtract.

Eeh, Riemann sums give the rationale for many numerical integration schemes..

 

[mathwonk]

There are two ways to compute an integral "exactly". Either take the actual limit of the Riemann sums, which is usually quite hard, or do as Halls emphasizes is much more usual, namely find an antiderivative. (His "always" means 99.9% of the time.)

If the antiderivative is elusive or totally unavailable, but usually not otherwise, one may use actual explicit Riemann sums to approximate the integral, as arildno says.

I.e. in almost all textbook problems, like integrating x^18, or cos^4(x) and so on, do as Halls said to get an "exact" answer. In the odd case where the integrand is 1/sqrt(x^5-x+1), or cos(x^2), do as arildno says to get an approximate answer.

I believe Halls thought you meant to get an exact answer, so his answer was chosen accordingly.

 

[HallsofIvy]

Eeh, Riemann sums give the rationale for many numerical integration schemes..

Yes, thank you. I didn't think of that. However, I would be inclined to say that most methods, such as Simpson's rule, can be interpreted as approximating the function, over sections of the interval rather than Riemann sums.

 

[lurflurf]

^The initially surprising thing is that approximating a function over an interval by an interpolating polynomial is equal to a Riemann sum.

 

[clm222]

correct me if I'm wrong (I've just started definite integration a few days ago): given an integral

\int_a^b f(x) dx = F(b) - F(a)

Also, what If I am trying to find the area under a curve, when the curve is completely random, and is based on no function.

Lastly: do i have to do anything special when some where from point a to point b, there is a negative area.

Thanks

 

[Muphrid]

You don't have to do anything special about the function taking on negative values. This is accounted for by the antiderivative.

"based on no function" makes no sense. If you mean you can't write the function in terms of known elementary functions, that's one thing, but I assume that this curve is still well-behaved--continuous, for instance. Just because you can't write it in terms of polynomials, sines, cosines, or anything like that doesn't make it not a function.

 

[clm222]

can you give me an example please?

 

[clm222]

wait a second, using

\int_a^b f(x) dx = F(b)-F(a)

I don't even need to know the function, just the values at point a and point b, am i correct?

One more question: given a function, say: f(x)=x²
to find the area under a curve (a to b) would I need to differentiate:

\int_a^b 2x dx

or would I simply plug in the function?

\int_a^b x² dx

 

[arildno]

However, I would be inclined to say that most methods, such as Simpson's rule, can be interpreted as approximating the function, over sections of the interval rather than Riemann sums.

True enough.
The calculative efficiency of the "strict" Riemann sum is low, relative to more advanced schemes that incorporate function approximation over the intervals as well.
(Of course, Riemann sums ALSO "approximate" functions over intervals, but usually only well when the intervals are extremely tiny)

The force of Riemann sums, as I see it, is the ease by which proofs in full generality may be given by means of them.

They are a bit like Cramer's Rule, dreadful as a computational tool, but optimal for some basic theoretical problems of how to prove something.

 

[micromass]

The force of Riemann sums, as I see it, is the ease by which proofs in full generality may be given by means of them.

No, Riemann sums are a horrible thing to use in proofs. Many proofs turn out to be very technical and ugly. Proving things like Fubini's theorem is (in my opinion) dreadful. Furthermore, Riemann sums are very restrictive things. Many nice results don't hold for them.

The good thing about Riemann sums is that the definition is easy and intuitive, but that's basically it.

 

[clm222]

If I could ask about three questions:

1: if I want to calculate the definite integral of say

f(x)=x²

would i need to differentiate before i find the integral or would i just plug in my function into the integral?

\int_a^b x² dx

or

\int_a^b 2x dx

2: when I use certain rules such as the trapazoid rule:

\int_a^b f(x) dx = (b-a) \frac{f(a)+f(b)}{2}

does f(x) equal the derivative of F(x) (given that F'(x)=F(x)), or the function value?

 

[HallsofIvy]

wait a second, using

\int_a^b f(x) dx = F(b)-F(a)

I don't even need to know the function, just the values at point a and point b, am i correct?

One more question: given a function, say: f(x)=x²
to find the area under a curve (a to b) would I need to differentiate:

\int_a^b 2x dx

What you are doing is integrating not differentiating.

or would I simply plug in the function?

\int_a^b x² dx

After integrating, you would not still use the integral sign- that's been done. What you are doing is evaluating x^2.
\int_a^b 2x dx= \left[x^2\right]_a^b= b^2- a^2

 

[clm222]

Ok, and that is the area of the function?

 

[HallsofIvy]

A "function" is not a geometric object and so does not have an "area". Do you mean the area below a graph of a function?

 

[clm222]

Yes, I thought that was one of the main points of definite integration?

 

[Muphrid]

Careful, I think you confused HallsOfIvy. I'm pretty sure you want to integrate x^2, do you not? There's no differentiation involved. You just do \int_a^b x^2 \; dx.

You can find the area beneath a graph using a definite integral, but that is not the definition of a definite integral. This is a frequent point of confusion when people try to integrate, say, functions in polar coordinates, where instead of approximating an area as a series of rectangles, one has to use triangles instead. The integral is a tool, but the form that integral takes depends on the geometry of the problem.

 

[clm222]

I'm not sure if this is right but, let's say I wanted to find the total acceleration from t₁ to t₂.

I could integrate the velocity from t₁ to t₂? v(t₂)-v(t₁)

but if i wanted to find the total work done finding the area under the graph of force from t₁ to t₂, would give the total work, woulnt it?

because using F(b)-F(a) would give me Δx(?)

and finding the area under a graph would give me the average value multiplied by time(?)

 

[Muphrid]

Total acceleration? What does that mean?

Integrating velocity does not give v(t_2) - v(t_1). It gives you \int_{t_1}^{t_2} v(t) \; dt = x(t_2) - x(t_1), the change in displacement.

Work is what you get when you integrate force with respect to position. Impulse (change in momentum) is what you get when you integrate force with respect to time.

As HallsofIvy said, I would quickly get out of this notion about areas. It is true, yes, for plotting functions on a rectangular grid, but that's not the core of what you're doing. You're just finding the cumulative effect of these functions over an interval, and when you're no longer plotting functions on a rectangular grid, you will get yourself very confused.

 

[Robert1986]

wait a second, using

\int_a^b f(x) dx = F(b)-F(a)

I don't even need to know the function, just the values at point a and point b, am i correct?

I don't think anyone has answered this part.

No, you can not do what you are saying because you have to know what is happening to f in (a,b). For example, the functions defined by:
<br /> f(x) = 1, x = 0, 1<br />
<br /> f(x) = 0, x \neq 0<br />

and
<br /> g(x) = 1<br />

are the same at a and b but their integrals are different.

 

[islam ana]

Ok, and that is the area of the function?

yes, it is the area of the function.
but this when the graph of the function lies on the first or second quadrent .

 

[clm222]

ok, thank you all very much for your patience and help, i have a clearer idea of definite integration.

 

[HallsofIvy]

Careful, I think you confused HallsOfIvy.

Yeah, I confuse easily!