There are set formulas that will help you find the antiderivative of your function. Many of these are listed under integrals in my reference tables. Integrals, by the way, are basically antiderivatives, set into a formula designed to tell you to take the antiderivative. So when I say take the integral I mean antiderivative, except for a whole function. And you will always have to add a " + C " afterward, because every integral has an unknown constant added to the equation. I will explain this shortly.
The antiderivative of xn is xn+1/(n+1). (Ex: antiderivative of x5 is x6/6) This will work for every value of n except for –1, because there n+1 will equal 0. The antiderivative of x-1 is ln x. I'm sure you recall that the derivative of ln x is 1/x, or x-1. If you accepted that, you can accept this. Though, to be honest with you, I have no idea why this is. It's frustrating.
The form for an integral is ∫ f(x) dx. When you solve the integral you remove the ∫ and the dx, and you are left with a function F(x), which is the antiderivative. Notice that I used capital F; I'm not sure if this is standard, but I've definitely seen it. I don't think I'll necessarily use this notation, but you should be aware of it.
y = x2 + 5x + 6 ==> y’ = 2x + 5 (this much you know)
I will now take the integral of y’ and see if it is correct in bringing back the original function.
∫ 2x + 5 dx = 2x1+1/(1+1) + 5x0+1/(0+1) = x2 + 5x yay!!! (got a little too excited :)
So you see that we were only part successful in bringing
back the original function. We are still missing a 6. It is for this reason that
every integral has a constant C added to it. The problem is that if there is any
number after a function, and you take the derivative, that number is lost.
When you want to integrate back, there is no way to know what the number was. So
we call it C. C can be zero, if there was no number standing alone. The actual answer
to our problem would look like x2
+ 5x + C. This C tells us that our original equation was the derivative of this equation,
whatever C is, so long as C is a constant (no x's in C).
Some rules for integration:
1) If you have an equation with parts separated by pluses
and minuses, you can break them up into separate integrals.
2) If you have an equation with parts separated by
multiplication or division, you can NOT break them up into separate integrals.
3) You may safely ignore the last rule if you are dealing
with a constant as one of the two parts. (as opposed to a variable)
*NOTE - The example for the
second rule uses the Substitution method, which is taught later in this page. Be
∫ 7x dx = 7∫ x dx = 7x2/2 + C Note that you can take out a constant if it factors through every term in the integral.
∫ 7x2 dx = 7x3/3 + C
∫ 7 dx = 7x + C
∫ 7x1/2 dx = 7x3/2/(3/2) +C = 14x3/2/3 + C For this one I had to flip the bottom fraction and multiply, to simplify it.
∫7x -4 dx = 7x-3/-3 + C
∫7x-3.143 dx = 7x-2.143/-2.143 + C
∫7x-1 dx = 7 ln |x| + C
∫7x-1/2 dx = 7x1/2/(1/2) + C = 14x1/2 + C
∫ sin x dx = -cos x + C I know this because I think in reverse: What would have a derivative of sin x? If cos x will give me -sin x, then I just negate both sides.
∫ sec2 x dx = tan x + C Again I simply think in reverse. There is no other way to solve this. Except memorization.
Important: Note that x1/2 can also be represented as √x, but it’s much easier to solve if everything is converted to a standard format. I also wrote 1/x4 as x-4 because once again it goes into the standard format and it is much easier to solve.
Integrals in particle movement
There are cases where you are given the velocity of a ball, and you are asked to find the displacement, or even given the acceleration. In these questions you will have to work backward through integrals to solve. The key to these problems is knowing the the derivative of displacement (overall distance traveled) is velocity, and the derivative of velocity is acceleration. This should be obvious, as you know that a derivative of an equation is its slope, or how quickly it is changing. Velocity is how quickly the displacement is changing, and acceleration is how quickly the velocity is changing! Here is a problem:
The equation for the velocity of an object is f(x) = x2 – 2. After 3 seconds, the object is 5 feet away. How far is the object after 4 seconds?
In order to solve this you must take the integral.
∫ x2 – 2 dx = x3/3 – 2x + C
Now I have an equation for displacement, but it’s not complete, so we can’t just plug 4 into it. I don’t yet know what the C represents. For this we must use the information given in the problem. If the object is 5 feet away after 3 seconds, we can set the equation equal to 5, and plug 3 into x, and see what the constant is. (Remember: in these equations, x always represents time, and the y-axis is all the other things)
5 = 27/3 – 6 + C C = 2
Now we can complete the equation:
D = x3/3 – 2x + 2
And we can solve for displacement when time (x) is 4.
D = 64/3 – 8 + 2 = 15.333
Integrals that need substitution can be fiendishly difficult. Not for me, maybe, but for most people – just a bit. The simpler substitution rules are taught in Calculus 1. I’m not sure why, since integrals are primarily in Calculus 2, but perhaps they just want you to have an introduction before you get buried in it.
The substitution that you need will only become easy with practice. But there are definite rules that you should work with.
∫ (x + 2)2 dx This problem is actually not that hard, because you can foil the equation and have 3 easy terms. But let’s try to solve it in its current state. The first thing you must do is find what to substitute. This will be very similar to chain rule problems. In this function, the inside function is x + 2. If we were to replace it with u, then the function u2 has an easy integral: u3/3. But we can’t just replace. The problem that comes up is that now we will be taking the integral of a function with u, and the dx at the end says take the integral of a function with x. The rule here is that dx = du/u’. So in the equation, replace the dx with a du divided by the derivative of you and then you can solve. After you have solved, though, remember to convert the u back to x.
∫ (x + 2)2 dx
Did you follow that progression? The du was divided by 1 because u’ (The derivate of u, or x+2) is 1. In the next example, the derivative will be something else.
∫ 7(3x + 2)3 dx = ∫ 7u3 du/3 = 7u4/(4 * 3) + C = 7(3x + 2)4/12 + C
This last integral may have been easier if you had used the rule that any constant that the entire inside of the integral is multiplied by can be moved to the outside of the integral. I will redo the last one using this.
∫ 7(3x + 2)3 dx = 7 ∫ u3 du/3 = 7 * u4/(4 * 3) + C = 7(3x + 2)4/12 + C
∫ (8x)-1 dx = ∫ u-1 du/8 = (1/8) ∫ u-1 dx = (1/8) ln |u| + C = ln |8x|/8 + C
∫ ex dx = ex + C
∫ e3x dx = ∫ eu du/3 = eu/3 + C = e3x/3 + C In this one I had to sub the 3x for a u.
The next stage of development is the integral is where you make a substitution that cancels out with another part of the integral allowing you to solve the problem. Notice what I wrote a couple of paragraphs up, that the dx is replaced by du/u'. In all of the previous problems, the inner function that I designated u was a simple function with a derivative of 1 or 3, etc., making the integral simple. In the following problems, the derivative of the inner function is more complex, but it will cancel.
∫ (2x + 3)(x2 + 3x
– 6)3 dx
The key to that equation is the understanding that you can only take an integral if there is one type of variable (such as u) inside the function you integrate, and that variable has to be the one referred to at the end, as in dx or du. In this equation, you created a du, u' equals (2x + 3), and all the x’s were inadvertently canceled out. If I had ended up with an equation that had both u's and x's inside, I could not solve it. Also, it would be illegal to breakup the two parts of this integral into two integrals. See my second rule above. Here’s a similar one:
∫ (2x + 3)(2x2 + 6x – 6)3 dx = ∫ (2x + 3)u3 du/(4x + 6) = ∫ u3 du/2 = u4/8 + C = (2x2 + 6x – 6)4/8 + C
In that one you are left with a 2 under the du, which was pulled out of the (2x + 3).
∫ x(ex)2x dx = ∫ xeu du/4x = ∫ eu du/4 = eu/4 + C = e2x * x/4 + C
I’m sorry in this last one I wrote the exponent of e in a funny form, but regular text doesn’t support exponents inside an exponent. I meant the exponent to be 2x2. But you get the idea.
∫ sin x dx = -cos x + C (This should be obvious, as the derivative of -cos is sin.
∫ sin 7x dx = (-cos 7x)/7 + C (This uses substitution of u = 7x, u' = 7)
∫ 7sin 7x dx = (-7 cos 7x)/7 = - cos 7x + C
∫ cos x sin x dx
This last one is a bit more complicated. Since I knew that I could cancel out one of them by dividing by the derivative of the other, I made one of them u, and then I was forced to divide the du by its derivative, canceling the cos x.
Note that many of these problems where the substitution cancels are very contrived, and designed to perfectly cancel out. There are more difficult problems, but you do not need to know them in Calculus 1.
Definite Integrals for area problems
This is the last bit of Calculus 1 I am going to discuss. It isn’t very complicated, if you can do integrals.
Basically, think of a curve that is passing along, and always above the x-axis. Underneath that curve I can describe an area. An area must have 4 bounds: the top and bottom, left and right. The top will be the curve. Unless I specify otherwise, the bottom will be the x-axis. The left and right I will always specify as values of x, and by them I mean vertical lines at those values. What is the area in this region? The way we find it is by taking the integral of the function that the curve represents. Then we plug in the two values of x designated, and subtract the side on the left from the side on the right. Previously, when we did an integral, we would not plug any values into the resulting equation.
These integrals are thought of as definite integrals, because they generally have two sides. Other integrals I did before did not have two sides set, and they were indefinite integrals. In definite integrals, since the C is the same on both sides, and we are subtracting one side from the other, we don’t even care what C is. We ignore it.
I will now proceed and give an example of how to solve an indefinite integral.
∫32 x + 6 dx
If I want to find the area beneath the curve (y = x + 6) from x = 2 to x = 3, I must do this integral. (Alternately, you could form a square and a triangle, and the areas are obvious.) Carefully note the syntax of the integral. Inside, we put the equation f(x) of the curve, as usual. Along the integral sign, at the top is right boundary of the region, and on the bottom is the left boundary of the region. Bigger over smaller.
∫32 x + 6 dx = [x2/2
= (32/2 + 6(3)) – (22/2 + 6(2)) = 8.5 Notice that unlike indefinite integrals, we have come up with an exact value. This is the area inside our region.
Here's another problem:
What is the area underneath the curve
-x^2 + 10 between -3 and 3?
The process of taking an integral to get the area underneath a curve is specific to when the part of the curve we are dealing with is always above the x-axis. The process must be modified if you want to deal with areas underneath. A more general formula will solve this problem, and also problems in which another curve forms the second boundary, as opposed to the x-axis.
Basically, when you take an integral, you always take the
top boundary minus the bottom boundary, inside your integral. In the problems we
have dealt with up until now, the bottom was the curve y = 0, so we took the
integral ∫ f(x) dx, which is the same as ∫ f(x) - 0 dx. (Top minus
Bottom) If the curve
is now entirely under the x-axis, and you want the area between the curve and
the x-axis, you take
And what if you want to take the area between two curves? Use the formula I just mentioned: Integral of the top curve minus the bottom curve. For instance, if you must take the integral between the line y = x and y = x2 between x equals 2 and 3, you will do ∫32 x2 - x dx. Solve it yourself. :) Also, if you are asked for the area between two curves over a region where they cross each other, you must break the integral into two parts, so that you are always doing the upper minus lower curve.That’s all folks! If this was too brief, look at the problem sets, or email me. Sign in the Guestbook, if you like!