Integral of (xsin(t))....Two Variables in Single Variable Calc Integral

In summary, the conversation discusses the evaluation of the expression d/dx ∫(limits: 0,x) x*sin(t) and suggests using calc-1 techniques. It is then determined that allowing x to be a function of t is allowed, and the product rule is used to evaluate the expression. The conversation also mentions the incorrect relevant equation and the use of integration by parts. It is concluded that proceeding with the product rule is acceptable, as x is treated as a constant in the integral.
  • #1
The Head
144
2
Homework Statement
Evaluate d/dx ∫(limits: 0,x) x*sin(t)
Relevant Equations
d/dx ∫ (0,x) f(t)= f(x)*x'

uv-∫vdu = ∫udv
I was told this problem could simply be solved with calc-1 techniques, so I'm tempted to say we could do d/dx(x∫(limits: 0,x) sin(t) dt. Then it's a simple product rule: d/dx (x) * ∫(0,x) sint dt + x * d/dx(∫ (0,x) sin (t) dt) = 1 - cos(x) + x*sin(x). However, I wonder if we have to allow that x is a function of t and then I would be forbidden from bringing x out of the integral. Is this allowed?

I tried moving the d/dx inside the integral too (∫ d/dx (x*sint) dt = ∫ sint dt + ∫x*d/dx(sint) dt.
∫ sint dt = 1 - cos(x)
∫x*d/dx(sint) dt means integration by parts (not really calc 1).

u=x
du=dx/dt
dv= d/dx(sint)
v =∫ d/dx (sint), but now the variables are getting all screwed up because the integral is with respect to t, but I'm taking d/dx. Was my first approach legal enough?
 
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  • #2
The Head said:
Homework Statement:: Evaluate d/dx ∫(limits: 0,x) x*sin(t)
Relevant Equations:: d/dx ∫ (0,x) f(t)= f(x)*x'
Your relevant equation is incorrect. Here's what it should say.
$$\frac d {dx}\int_0^x f(t) dt = f(x)$$
Notice that I have included dt in the integrand, which you don't show in the homework statement. Notice also that x' is not part of the answer.
This equation is a consequence of the first part of the Fundamental Theorem of Calculus.

I'm guessing that the full Homework Statement indicates that you are to evaluate this expression:
$$\frac d {dx}\int_0^x x\sin(t) dt$$
I'm guessing, because your version doesn't indicate which is the variable of integration, so I'm uncertain whether the last thing in the integral should be dt or dx.
The Head said:
uv-∫vdu = ∫udv

I was told this problem could simply be solved with calc-1 techniques, so I'm tempted to say we could do d/dx(x∫(limits: 0,x) sin(t) dt. Then it's a simple product rule: d/dx (x) * ∫(0,x) sint dt + x * d/dx(∫ (0,x) sin (t) dt) = 1 - cos(x) + x*sin(x). However, I wonder if we have to allow that x is a function of t and then I would be forbidden from bringing x out of the integral. Is this allowed?

I tried moving the d/dx inside the integral too (∫ d/dx (x*sint) dt = ∫ sint dt + ∫x*d/dx(sint) dt.
∫ sint dt = 1 - cos(x)
∫x*d/dx(sint) dt means integration by parts (not really calc 1).

u=x
du=dx/dt
dv= d/dx(sint)
v =∫ d/dx (sint), but now the variables are getting all screwed up because the integral is with respect to t, but I'm taking d/dx. Was my first approach legal enough?
 
  • #3
Mark44 said:
Your relevant equation is incorrect. Here's what it should say.
$$\frac d {dx}\int_0^x f(t) dt = f(x)$$
Notice that I have included dt in the integrand, which you don't show in the homework statement. Notice also that x' is not part of the answer.
This equation is a consequence of the first part of the Fundamental Theorem of Calculus.

I'm guessing that the full Homework Statement indicates that you are to evaluate this expression:
$$\frac d {dx}\int_0^x x\sin(t) dt$$
I'm guessing, because your version doesn't indicate which is the variable of integration, so I'm uncertain whether the last thing in the integral should be dt or dx.

Yes, $$\frac d {dx}\int_0^x x\sin(t) dt$$ is correct. Thanks, apologies on forgetting the dt. Is it OK to then just proceed with $$\frac d {dx}x\int_0^x sin(t) dt$$ and try for the product rule? It doesn't seem to specify if x is a function of t, but given how involved it gets when I don't make this assumption, I feel like this must be allowed.
 
  • #4
The Head said:
Yes, $$\frac d {dx}\int_0^x x\sin(t) dt$$ is correct. Thanks, apologies on forgetting the dt. Is it OK to then just proceed with $$\frac d {dx}x\int_0^x sin(t) dt$$ and try for the product rule? It doesn't seem to specify if x is a function of t, but given how involved it gets when I don't make this assumption, I feel like this must be allowed.
That seems OK to me. I can't come up with a good reason why it isn't.
If we just look at the integral, it would be
$$\int_{t=0}^{t=x} x\sin(t)dt$$
As far as the integration is concerned, x is just a constant.

BTW, I appreciate you efforts in using LaTeX. It makes things much easier to read.
 
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Related to Integral of (xsin(t))....Two Variables in Single Variable Calc Integral

1. What is the formula for finding the integral of (xsin(t)) in single variable calculus?

The formula for finding the integral of (xsin(t)) in single variable calculus is ∫xsin(t) dt = -xcos(t) + sin(t) + C, where C is the constant of integration.

2. How do you solve for the integral of (xsin(t)) using substitution?

To solve for the integral of (xsin(t)) using substitution, let u = sin(t) and du = cos(t) dt. Then, the integral becomes ∫xsin(t) dt = ∫xu du = (x/2)u^2 + C = (x/2)sin^2(t) + C.

3. What is the importance of the constant of integration in the integral of (xsin(t))?

The constant of integration is important because it represents all possible solutions to the integral. In other words, the constant of integration accounts for any unknown or missing information in the original function.

4. Can the integral of (xsin(t)) be solved using integration by parts?

Yes, the integral of (xsin(t)) can be solved using integration by parts. Let u = x and dv = sin(t) dt. Then, du = dx and v = -cos(t). The integral becomes ∫xsin(t) dt = -xcos(t) + ∫cos(t) dx = -xcos(t) + sin(t) + C.

5. What are some real-world applications of the integral of (xsin(t))?

The integral of (xsin(t)) has various real-world applications, such as calculating the work done by a force acting on a particle moving along a curved path, determining the displacement of a pendulum, and finding the average value of a periodic function over a given interval.

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