Solve Quick Integral Question with Step-by-Step Solution | Attached Picture

[unscientific]

Homework Statement

The question is as attached in the picture.

The Attempt at a Solution

I haven't got to the answer, but I wonder if this is OK:

g(y) = ∫ f(x,y) dx

= F(y) - F(0,y)

(dg/dy) = f(y) - f(0,y)


Since the 'x' is just a dummy variable where the final function g will be in terms of y. By integrating it with respect to x, then filling it in with 0 and y, by differentiating g does it reverse back the integration process and simply filling it with 0 and y?

 

[LCKurtz]

Homework Statement

The question is as attached in the picture.

I don't see any question in that picture.

 

[unscientific]

I don't see any question in that picture.

Sorry that was just part of the question, I am trying to find out if its OK to do the steps above.

 

[LCKurtz]

Homework Statement

The question is as attached in the picture.

The Attempt at a Solution

I haven't got to the answer, but I wonder if this is OK:

g(y) = ∫ f(x,y) dx

= F(y) - F(0,y)

(dg/dy) = f(y) - f(0,y)


Since the 'x' is just a dummy variable where the final function g will be in terms of y. By integrating it with respect to x, then filling it in with 0 and y, by differentiating g does it reverse back the integration process and simply filling it with 0 and y?

I can't follow this. $g(y) = \int f(x,y)\, dx$ doesn't make any sense to me. It's an indefinite integral and surely there would be an $x$ on the left side. Then you have $F(y)-F(0,y)$. I gather that is supposed to represent an antiderivative, but is $F$ a function of one or two variables? Then in your last line you have $f(y) - f(0,y)$. But $f$ is a function of two variables, so $f(y)$ is meaningless.

Anyway, whatever it is you are trying to do, try it with $f(x,y) = y^2\cos(x)$ or something like that. It has to make sense and work for something simple to have any chance of being true in general.

 

[unscientific]

I can't follow this. $g(y) = \int f(x,y)\, dx$ doesn't make any sense to me. It's an indefinite integral and surely there would be an $x$ on the left side. Then you have $F(y)-F(0,y)$. I gather that is supposed to represent an antiderivative, but is $F$ a function of one or two variables? Then in your last line you have $f(y) - f(0,y)$. But $f$ is a function of two variables, so $f(y)$ is meaningless.

Anyway, whatever it is you are trying to do, try it with $f(x,y) = y^2\cos(x)$ or something like that. It has to make sense and work for something simple to have any chance of being true in general.

Sorry the integral is meant to be a definite integral as in the question, I'm not sure how to use latex...

How about if

g(y) = ∫ f(x,y) dx (definite integral)

= F(y,y) - F(0,y)

then

dg/dy = f(y,y) - f(0,y)

Does this make more sense?

 

[Psychosmurf]

You have the right idea, but the steps you're using are wrong. The function in the integral is not a function of x and y, it is solely a function of x, so your antiderivatives in the second part don't make any sense.

 

[Psychosmurf]

Sorry the integral is meant to be a definite integral as in the question, I'm not sure how to use latex...

How about if

g(y) = ∫ f(x,y) dx (definite integral)

= F(y,y) - F(0,y)

then

dg/dy = f(y,y) - f(0,y)

Does this make more sense?

Integrate the function in the integral with respect to x while keeping y constant.

 

[unscientific]

Integrate the function in the integral with respect to x while keeping y constant.

Yes that's right but are my steps right? I'm concerned about the conversion from 'F' to 'f' by differentiation. Since they are entirely in terms of y after integration.

 

[Ray Vickson]

Sorry the integral is meant to be a definite integral as in the question, I'm not sure how to use latex...

How about if

g(y) = ∫ f(x,y) dx (definite integral)

= F(y,y) - F(0,y)

then

dg/dy = f(y,y) - f(0,y)

Does this make more sense?

You don't need to use LaTeX (although some posters might complain if you don't), but you DO need to write things explicitly. For example, if you want to write
$$\int_a^b f(x,y) \, dx$$ in plain text you can just write int( f(x,y) dx, x=a..b), for example.

RGV

 

[voko]

Lleibniz integral rule:

$$\frac {d} {dy} \int_{a(y)}^{b(y)} f(x, y)dx = \frac {db(y)} {dy} f(b(y), y) - \frac {da(y)} {dy} f(a(y), y) + \int_{a(y)}^{b(y)} \frac {\partial} {\partial y} f(x, y)dx$$

 

[Psychosmurf]

Lleibniz integral rule:

$$\frac {d} {dy} \int_{a(y)}^{b(y)} f(x, y)dx = \frac {db(y)} {dy} f(b(y), y) - \frac {da(y)} {dy} f(a(y), y) + \int_{a(y)}^{b(y)} \frac {\partial} {\partial y} f(x, y)dx$$

Would it apply, because the function isn't continuous over [0,y]?

 

[Psychosmurf]

OP, try integrating by parts.

EDIT: It worked for me. Integrate by parts and then fundamental theorem of calculus combined with the hint that f(0) = 0.

 

[voko]

OP, try integrating by parts.

And then the Leibniz rule :)