# Integration Problem I Cannot Figure Out

1. Nov 29, 2004

### mathemagician

My Professor in my calculus class (1st year) left us with this question at the end of lecture today and told us to think about it. I am baffled as to how to solve it. Anyways, here is what he gave us.

$$\int_{x}^{xy} f(t) dt$$
This is independent of x.

If $$f(2) = 2$$, compute the value of
$$A(x) = \int_{1}^{x} f(t)dt$$

for all $$x > 0$$

He then gave us a hint saying since it is independent of x, the function will be in terms of y.
$$g(y) = \int_{x}^{xy}f(t)dt$$
He also told us the final answer is $$4 \ln x$$

Does this make any sense? I would appreciate it if someone can show me how to solve this.

2. Nov 29, 2004

### arildno

To solve this, differentiate g(y) with respect to x:
$$0=\frac{d}{dx}g(y)=yf(xy)-f(x)\to{f}(xy)=\frac{f(x)}{y}$$
then, differentiate g(y) with respect to y:
$$\frac{dg}{dy}=xf(xy)=\frac{xf(x)}{y}$$

Hope this helps..

3. Nov 29, 2004

### mathemagician

I am a little confused after spending an hour thinking about it. But I think I have something.

Since $$f(2) = 2$$ then $$\frac{dg}{dy} = \frac{2f(2)}{y} = \frac{4}{y}$$

Then we can replace $$f(t)$$ with $$\frac{4}{y}$$

Going back we can now solve for $$A(y) = \int_{1}^{x} \frac{4}{y} dy = 4 \int_{1}^{x} \frac{1}{y} dy = 4[\ln |x| - \ln (1)]$$
and since $$x > 0$$ we finally have:

$$A(y) = 4 \ln x$$

OK, so is this right? I'm a little bit troubled with doing the substitution of f(2) = 2, can you explain to me how that might be justified?

I also have a question about your hint, arildno. Just the first line.

how is it possible that you set $$\frac{d}{dx}g(y) = 0$$? And could you explain $$yf(xy) - f(x)$$ where that came from?

Thanks

4. Nov 29, 2004

### arildno

1) g is solely a function of the variable "y".
Hence, differentiating it with respect to some other variable it does not depend on, yields zero.
2) Using the Leibniz rule for differentiating an integral where the bounds depend on your variable, reads:
$$\frac{d}{dx}\int_{x}^{xy}f(t)dt=f(xy)\frac{d}{dx}xy-f(x)\frac{d}{dx}x=yf(xy)-f(x)$$

3. Since g(y) is independent of x, so is $$\frac{dg}{dy}$$
Hence, we must have:
xf(x)=K, where K is some constant.
We can determine K, with noting 2f(2)=4, that, is,
xf(x)=4 (implying f(x)=\frac{4}{x}), or $$\frac{dg}{dy}=\frac{4}{y}=f(y)$$

5. Nov 29, 2004

### mathemagician

Thank you. I understand. Its much clearer now.

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