Category Archives: Calculus

Integration Revision - Integration By Substitution

Now, looking at the other integral from Oct 13th...


In my opinion, this was scarier than the first, but looking at it again, what makes it scary is the various powers. e to the power of x to the power of 2 and such.

But why should this be scary? There's practically only one tool that one could use to solve this: the only tool that includes functions of functions, integration by substitution:

\int f(g(x))\:g^{\prime}(x)\:dx=\int f(u)\: du

In our case, the inner function:

and the outer function:

Plugging all these into our "integration by substitution" tool gives:

\int -4xe^{-2x^{2}}\:dx

But this is slightly different to the original integral due to that minus sign. Of course this is trivial to deal with as:

\int^{x}_{0}4xe^{-2x^{2}}\:dx=-\int^{x}_{0} -4xe^{-2x^{2}}\:dx

Great! So now we have the structure we require to apply integration by substitution! We can substitute all of that with f(u), and all the scary bits go away. So:


=-\int^{x}_{0} -4xe^{-2x^{2}}\:dx

=-\int^{u}_{0} e^{u}\:du

=-[ e^{u}]^{u}_{0}



Then substituting u (=g(x)) back in:


Integration Revision - Integration By Parts, Twice

So looking at the first one from a few days ago (Oct 13th):

\int^{\infty}_{0}x^{2}\:\lambda\: e^{-\lambda x}\:dx=\frac{2}{\lambda^{2}}

I was totally lost with this. But running through some old tricks made this seem a lot more approachable. First off, moving the constant out brings a bit more clarity to the integrand,

\int^{\infty}_{0}x^{2}\lambda e^{-\lambda x}\:dx=\lambda\int^{\infty}_{0}x^{2} e^{-\lambda x}\:dx

But given the x^{2} and the exponential, we'll also probably need to use integration by parts:

\int^{b}_{a} f(x)g^{\prime}(x)dx=[f(x)g(x)]^{b}_{a}-\int^{b}_{a} f^{\prime}(x)g(x)dx

So, as above, letting:
g^{\prime}(x)=e^{-\lambda x},

we have:
g(x)=-\frac{1}{\lambda}e^{-\lambda x}

Then plugging f, f^{\prime}, g and g^{\prime} into our lovely integration by parts tool above gives:

\lambda\int^{\infty}_{0}x^{2}\: e^{-\lambda x}\:dx

=\lambda\left(\left[-x^{2}\: \frac{1}{\lambda}\:e^{-\lambda x}\right]^{\infty}_{0}-\int^{\infty}_{0}-2x\:\frac{1}{\lambda}\: e^{-\lambda x}\:dx\right)

=\lambda\left(0-0+2\frac{1}{\lambda}\int^{\infty}_{0}x\: e^{-\lambda x}\:dx\right)

=2\int^{\infty}_{0}x\: e^{-\lambda x}\:dx

Which looks very familiar with what we've started with, except x^{2} is now just x! If we can reduce x by another power, we'll end up with just 1 which will surely give us a much easier integral to solve.

So applying integration by parts again, but letting:
g^{\prime}(x)=e^{-\lambda x},

we have:
g(x)=-\frac{1}{\lambda}e^{-\lambda x}


2\int^{\infty}_{0}x\: e^{-\lambda x}\:dx

=2\left(\left[-x\: \frac{1}{\lambda}\:e^{-\lambda x}\right]^{\infty}_{0}-\int^{\infty}_{0}-\frac{1}{\lambda}\: e^{-\lambda x}\:dx\right)

=2\left(0-0+\frac{1}{\lambda}\int^{\infty}_{0} e^{-\lambda x}\:dx\right)

=\frac{2}{\lambda}\int^{\infty}_{0} e^{-\lambda x}\:dx

=\frac{2}{\lambda}\left[-\frac{1}{\lambda}\:e^{-\lambda x}\right]^{\infty}_{0}




Integration Revision

Putting these in a safe place for later. Found these to be a really good revision questions for integration.

\int^{\infty}_{0}x^{2}\lambda e^{-\lambda x}\:dx=\frac{2}{\lambda^{2}}



For context, the first one formed part of a question that required me to find the variance of a random variable, and the second one involved having to find the c.d.f. from a p.d.f.

(for my own reference, this was Book 1, Activity 5.2, p. 54 and Activity 6.1, p.62)