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

  • Thread starter The Head
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  • #1
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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?
 

Answers and Replies

  • #2
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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
141
2
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
35,138
6,889
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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