# Integration help

#### manyarrows

Hello, I have the following poblem:

$$\displaystyle v"=\frac{q}{6L}(L^3-3L^2x+3Lx^2-x^3)$$

Upon integrating I get:

$$\displaystyle v'=\frac{q}{6L}(L^3x-\frac{3L^2x^2}{2}+Lx^3-\frac{x^4}{4}+C_1)$$

The book derives v" as follows:

$$\displaystyle v"=\frac{q}{6L}(L-x)^3$$

And upon integration gets:

$$\displaystyle v'=\frac{q}{24L}L-x)^4+C_2$$

I follow how the text does the problem, but why is mine wrong? Shouldn't I be able to expand the $$\displaystyle (L-x)^3$$ and integrate and get the same answer. I thought that they might be the same if I set them equal (The nature of the problem allows for me to solve for $$\displaystyle C_1$$ and the book gives $$\displaystyle C_2$$. No luck.

Where did I screw up? Thank you.

#### BBAmp

You actually did it right but when you expand and integrate you will end up with a single term missing which you can actually find by setting the book answer and your expansion answer equal to each other. Let me explain:

From your answer: $$\displaystyle v'=\frac{q}{6L}(L^3x-\frac{3L^2x^2}{2}+Lx^3-\frac{x^4}{4} + C_{1})$$

You can leave the c out there since its an arbitrary constant

From the book: $$\displaystyle v'=-\frac{q}{24L}(L-x)^4+C_2$$ The book answer should have this negative sign on front. If you solve by u-substitution you will see that this is true.

when you expand $$\displaystyle -(L-x)^4$$ you get

$$\displaystyle -(L^4-4L^3x+6L^2x^2-4Lx^3 + x^4)$$

Set the book answer and yours equal to each other and move the constants $$\displaystyle C_{1}[/MATh] and \(\displaystyle C_{2}$$ to one side and combine them to make a new arbitrary constant $$\displaystyle C$$ after doing that you should get this equation

$$\displaystyle \frac{q}{6L}(L^3x-\frac{3L^2x^2}{2}+Lx^3-\frac{x^4}{4}) = -\frac{q}{24L}(L^4-4L^3x+6L^2x^2-4Lx^3 + x^4) + C$$

if you cancel the terms you will end up with $$\displaystyle -L^4 = C$$.

The answer for $$\displaystyle C_{1}$$ is not necessarily $$\displaystyle L^4$$ since $$\displaystyle C_{1} - C_{2} = C$$ but that is beside the point. The reason why you do not get $$\displaystyle L^4$$ in your answer is because it is a constant and integrating directly hides that constant after the integration in $$\displaystyle C_{1}$$. Does this make sense?

In u-substitution more or less all the variables are preserved regardless of what is being integrated.\)

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#### manyarrows

So in $$\displaystyle v'=\frac{q}{6L}(L^3x-\frac{3L^2x^2}{2}+Lx^3-\frac{x^4}{4}+C_1$$ you mean that when I solve for $$\displaystyle C_1$$ it should work out?

I tried this with the following info v'(0)=0 and I got $$\displaystyle C_1=0$$.
v'= the slope of the beam deflection curve and v'(0)=0 corresponds to a fixed anchor point.

Assuming I don't have the book answer to use as help, can this problem be done without factoring it out.

These are Mech of Material problems and having to factor will add an extra tedious step, especially on exams.

Thanks again

#### BBAmp

So in $$\displaystyle v'=\frac{q}{6L}(L^3x-\frac{3L^2x^2}{2}+Lx^3-\frac{x^4}{4}+C_1$$ you mean that when I solve for $$\displaystyle C_1$$ it should work out?

I tried this with the following info v'(0)=0 and I got $$\displaystyle C_1=0$$.
v'= the slope of the beam deflection curve and v'(0)=0 corresponds to a fixed anchor point.

Assuming I don't have the book answer to use as help, can this problem be done without factoring it out.

These are Mech of Material problems and having to factor will add an extra tedious step, especially on exams.

Thanks again
So I edited my post a few times mostly because I am weird like that but also to clarify a few things:

Technically your answer is right as it was originally because you are only missing a single term that is a constant with respect to x. If a teacher marks that wrong you can contest your answer by saying L^4 is zero when you take its derivative with respect to x so it does not matter if it is hidden in a constant C or not.

Also, the way I wrote it you aren't exactly solving for $$\displaystyle C_{1}$$ but rather for another constant that is a combination of $$\displaystyle C_{1}$$ and $$\displaystyle C_{2}$$. This is a minor nitpicky point but what I hoped to show you was that there was a missing term and it was hidden as an arbitrary constant.

The best way to solve these types of integration problems is to use u-substitution. Since you have $$\displaystyle (L-x)^3$$ with respect to $$\displaystyle x$$, using u-sub will get rid of the $$\displaystyle x$$ and you will end up with a simple varible $$\displaystyle u^3$$ with $$\displaystyle -du$$ (because if $$\displaystyle u = L - x$$ then $$\displaystyle du = -dx$$). Integrating $$\displaystyle -\int u^3du$$ is very easy.

Let me know if something doesn't make sense.

#### manyarrows

I see what you were showing me and I see how they are the same and I am happy to know I'm not way out in left field.

The problem I have is I need to solve for the constant C when I work these problems. They give you parameters based on the beam design that allow this. This problem fo instance is a cantilevered beam with a distributed load on it. The equation v' is the slope of the equation giving the bending curve of the beam. The beam is put on an xy grid with the left hand side at the origin. The origin corresponds to the fixed anchor which means there is no bending of the beam at that point thus v'(0)=0 and I should be able to set v'=0 and solve for C. But When I do this with my form I get $$\displaystyle C_1=0$$. The book finds $$\displaystyle C_2=-\frac{qL^3}{24}$$ which I follow when they do it.

I solved for $$\displaystyle C_1=\frac{24L^4+qL^3}{24}$$ using your explanation and cant figure out why I can't get that doing it from the expanded way.

Thanks for your time I really appreciate this.

#### BBAmp

Okay, I see. The terms are unfamiliar to me but I can certainly clarify the math.

If you choose to use your method in a similar problem, it is probably best to take out the C altogether so that it is alone. I believe this is the standard way to write constants. Let me show you what I mean and why this can help:

$$\displaystyle v' = \frac{q}{6L}(L^3x - \frac{3L^2x^2}{2} + Lx^3 - \frac{x^4}{4} + C_{1}) = \frac{q}{6L}(L^3x - \frac{3L^2x^2}{2} + Lx^3 - \frac{x^4}{4}) + C$$

because $$\displaystyle C_{1}$$ is an arbitrary constant and $$\displaystyle C_{1}(\frac{q}{6L})$$ becomes some other arbitrary constant $$\displaystyle C$$. You should also factor out a $$\displaystyle \frac{1}{4}$$ out of the polynomial so you get

$$\displaystyle \frac{q}{24L}(4L^3x - 6L^2x^2 + 4Lx^3 - x^4) + C$$.

Using this I believe you will be able to get the answer you need.

I guess if there is anything to take away from this it is that simpler methods should be used to solve problems that would otherwise be difficult using brute-force methods.

I was glad to help out.

#### BBAmp

hm wait a minute it doesn't.. let me think through this some more.

#### BBAmp

Okay I think I figured it out conceptually. We were integrating one variable but we still have to account for the other. If we do this we will end up with two constants, one for $$\displaystyle x$$ and one for $$\displaystyle L$$. The constant for $$\displaystyle L$$ stays in the polynomial part as x as a function of $$\displaystyle L$$ or $$\displaystyle x(L)$$ because we are deriving or integrating $$\displaystyle L$$ with respect to $$\displaystyle x$$ and the constant for $$\displaystyle x$$ is the constant that you leave outside the polynomial.

When you end up with two constants like this you will have to solve for one before solving the other. As to exactly how this is done I do not remember but I am fairly confident it involves using $$\displaystyle v''$$ to somehow solve for $$\displaystyle x(L)$$. I have only done this extensively in differential equations so I think the problem is best solved by u-substitution.

• manyarrows

#### manyarrows

Thanks and have a good day(Hi)

• BBAmp