OK, get ready for some Calculus-Fu!

We have now said that rather than taking pesky limits of Riemann sums to calculate areas under curves (ie. definite integrals), all we need is to find an antiderivative of the function that we are looking at.

As a reminder, to calculate the definite integral of a continuous function, we have:


\int_a^b f(x)dx=F(b)-F(a)


where F is any antiderivative of f

Remember that to calculate the area under the curve of f(x)=x^4 from, let’s say 2 to 5, we had to write:


\int_2^5 x^4 dx=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i)\Delta x=\lim_{n\rightarrow \infty} f\left(2+\frac{3i}{n}\right)\frac{3}{n}=\lim_{n\rightarrow\infty}\frac{3}{n}\left(2+\frac{3i}{n}\right)^4


And at that point we had barely even started because we still had to actually evaluate this sum, which is a hell of a calculation…then we have to calculate the limit. What a pain.


Now, we are told that all we have to do is to find any antiderivative of f(x)=x^4 and we are basically done.


Can we find a function which, when we take its derivative gives us x^4? Sure, it’s going to be something times x^5, but when you take the derivative of this it gives you 5x^4 so we have five times too much, so the answer is \frac{1}{5}x^5. Great. This is an antiderivative of x^4. Of course we could add on any constant, but we don’t need to. We are just asked to find ANY antiderivative. Now, using the FTC part 2, we have:


\int_2^5 x^4 dx=\frac{1}{5} 5^5-\frac{1}{5}2^5


No Riemann sums, no limits, no worrying about whether we are using the left point rule or the right point rule. Just find an antiderivative, and you are done. Amazing!


There’s a small caveat here…here’s the fine print…it’s not always easy to find the antiderivative of a function….so over the coming posts we are going to develop lots of skills to calculate antiderivatives of ever more complicated functions.


If you wanted to calculate the area under the cosine function for some arbitrary region it would be very very tricky! Now, we just need an antiderivative of cosine….which is sine. Boom!


\int_a^b \cos(x) dx=\sin(b)-\sin(a)


To get to this point though we needed to use the FTC part 1, which was used to prove the FTC part 2.  Now that we’ve got it, we’ll use it all over the place.


How clear is this post?