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I'm given that -> $$\frac{dy(x)}{dx}=x$$ I integrate both sides -> $$\int \frac{dy(x)}{dx}dx=\int x\,dx$$

Would it be correct if i canceled out the $dx$s and wrote -> $$\int dy=\int x\, dx$$

therefore

$$y= \frac{x^2}{2}+C,$$ where $C\in \mathbb{R}$

1likeMaths
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EldarRahim
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1 Answers1

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Yes, this would be fine. It's just like integration by substitution. $$\int y'(x) \,dx$$ Substitute $u=y(x)$. Then $du=y'(x)\,dx$, which is exactly equal to the integrand. So you simply get $$\int \,du\equiv\int\,dy$$

John Doe
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    +1 It is good to remember that substitution is what is really what is happening; the $dx$'s don't cancel out in the same way real numbers do. It just looks like it. – Plutoro Jan 23 '18 at 20:27
  • Yes understood, thanks. But don't we cancel out $dx$s when we went from $u=y(x)$ to $du=y'(x),dx$? The steps : $$\frac{du}{dx}=y'(x)$$ $$\require{cancel}\frac{du}{\cancel {dx}}\cancel{dx}=y'(x),dx$$ $$du=y'(x)dx$$ – EldarRahim Jan 23 '18 at 20:52
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    @EldarRahim Yes and no. We worked out $du$ as $du=d(y(x))$ which is "a small change in $y$. $$\begin{align}y(x+\delta x)&=y(x)+y'(x),\delta x +o(\delta x^2)\\implies dy=y(x+dx)-y(x)&=y'(x),dx\end{align}$$ But yes, you're right, it looks exactly like you are cancelling, and in most cases, you'll be fine to just "cancel" them as if they were regular numbers. This is one of the great benefits of the notation we use for derivatives. – John Doe Jan 24 '18 at 00:31
  • John Doe Sir, please check my edited question when your are free https://math.stackexchange.com/questions/2621182/integration-area-under-the-curve?noredirect=1#comment5413525_2621182 – Nikhil Kashyap Jan 26 '18 at 05:11