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My textbook says the area under a graph is given by: $\smallint ydx$

And it then goes on to say by the chain rule:

$$\smallint ydx = \smallint y{{dx} \over {dt}}dt$$

Could someone explain to me how this is reached? I can't seem to wrap my head around this. Thank you!

Américo Tavares
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seeker
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2 Answers2

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What you state is a formula for integration by substitution. The details of the proof of validity of this formula are usually left out on A-level courses. There is really no reason to be wishy-washy about this argument at all (other than for time constraints), so I thought I'd post a more rigorous answer. It goes beyond what you'll need to know for A-level, but should be understandable with a bit of work. Let me know if there's anything you want clarifying.

If $f:[a,b]\rightarrow\mathbb{R}$ is continuous, define $F(x)=\int_a^xf(u)du$ for $x \in [a,b]$. Let $g[\alpha,\beta]\rightarrow\mathbb{R}$ be continuously differentiable and such that $g(\alpha)=a,g(\beta)=b$.

We aim to show that $\int_a^bf(x)dx=\int_{\alpha}^{\beta}f(g(t))g'(t)dt$. This is seen to be the desired formula for integration by substitution with a little thought. (We can think of $g(t)$ as being the reparameterisation of $x$ and so "$g'(t)=\frac{dx}{dt}$".)

Let $h(t)=F(g(t))$. By the chain rule, $h'(t)=F'(g(t))g'(t)=f(g(t))g'(t)$. Thus:$$\int_\alpha^\beta f(g(t))g'(t)dt=\int_\alpha^\beta h'(t)dt$$ so$$\int_\alpha^\beta f(g(t))g'(t)dt=h(\beta)-h(\alpha)=F(g(\beta))-F(g(\alpha))=\int_{a}^{b}f(dx)dx$$

This argument still relies on some unproven facts, but to prove everything we need would require giving a proper definition of integration. Treating derivatives as fractions (e.g. $\frac{dx}{dt}dt=dx$) is often a good way to get an intuitive understanding of various theorems such as this and as such is frequently used to justify them, even though it isn't rigorous.

Tom Oldfield
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  • thank you for your answer, a more rigorous definition is precisely what I need. The thing is i'm quite lacking when it comes to reading math symbolically, could you explain this in simpler terms if possible? It would be very much appreciated, thank you. – seeker Apr 24 '13 at 22:23
  • @Assad No problem, which bits in particular? – Tom Oldfield Apr 24 '13 at 22:37
  • I feel embarrassed saying this but a lot of it is impenetrable to me, all the functional notation is confusing. Could you perhaps recommend me a better book to learn from? The edexcel books are far too short on detail for my liking. But if you are going to go ahead and explain this to me could you do so by explaining what each bit of functional notation you use means? Thank you for your help. – seeker Apr 24 '13 at 22:45
  • @Assad, There is no need to be embarrassed, this site is here to help! In terms of functional notation, think about what we mean when we say $\int ydx$. Here $y$ is actually a function of $x$. For example, if $y=x^2$, $\int ydx = \frac13x^3+c$ for some constant $c$. What I am doing here is writing $f(x)$ rather than $y$ to illustrate the fact that $y$ is a function of $x$. Saying $f:[a,b]\rightarrow\mathbb{R}$ means that $f$ is a function that accepts values of $x$ in the closed interval $[a,b]$ (The set of all points between $a$ and $b$ inclusive) and returns values in the real numbers. – Tom Oldfield Apr 24 '13 at 22:59
  • I also use, for example, $g'(t)$ to mean the function that is the derivative of $g$ with respect to $t$. The notation $f(g(t))$ means that we first apply the function $g$ to $t$ and then apply $f$ to the result. For example, if $f(x)=2x$ and $g(t)=t+1$ then $f(g(t))=f(t+1)=2(t+1)$, and also $g(f(t))=g(2t)=2t+1$. I hope this helps, let me know if you want some more help! – Tom Oldfield Apr 24 '13 at 23:03
  • Thank you that's cleared things up a bit, what do you mean F(x)= ∫f(u)du ? Where did the u come from? – seeker Apr 24 '13 at 23:06
  • @Assad In that case, $F(x)=\int_a^x f(u)du$, we call $u$ a dummy variable. This is because we are doing a definite integral, $u$ is "integrated out". Take for example $f(u) = u^2$. Then if $F(x) = \int_0^x f(u)du$, $F(x)=[\frac13 u^3]^x_0=\frac13 x^3. Since we substitute $x$ into the expression where $u$ appears, it is a "dummy", it doesn't do anything. In this expression, the $x$ dependence is in the limit of the integral, rather than in the integrand. Is that a little clearer? – Tom Oldfield Apr 24 '13 at 23:16
  • Tom thank you for your help, I'm going to call this a night, and use your notes to attack this tomorrow, thank you for all your help! – seeker Apr 25 '13 at 00:09
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$dx=dx \dfrac{dt}{dt}$ that is because $\dfrac{dt}{dt}=1$.So,

$$\int ydx = \int y{{dx} \over {dt}}dt$$

In case you do not no what chain rule is - Here is a link

Shaswata
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