DinS          Written on 2018/4/26

1. Techniques of integration

Using F.T.C. we’ve made our life easier. But we’re just changing the problem of working with limit to another problem of finding anti-derivative. As we know already, finding anti-derivative is not an easy task. Here we’ll discuss some commonly-used techniques to find anti-derivative.

(1) u-substitution

This is kind like chain rule in reverse.

Algorithm:

If we can make a function look like this:

We can let u = g(x), then . We can write it as .

Now here’s the magic: we can multiply both sides by dx and get du = g’(x)dx.

Thus we can write the original problem to

Wait a minute! This seems crazy and illogical. What happened? I’ll go into detail here.

If we just replace the symbol literally we get , which is f(u) + c.

Seems fishy. But the most crazy part is how we get du.

Can we really multiply dx on both sides? Isn’t dx supposed to be part of ?

If you make believe, you’ll find that du, dx played a critical role during the process. They aren’t just symbols. My suggestion is that take those as variables.

The algorithm seems totally fishy, but once you apply it in real problem you’ll see it really works.

It’s hard to find an anti-derivative for this. If we can’t then F.T.C. won’t help us at all. It’s time for u-substitution to shine.

The hard part is what should u be. This need flash of insight. As a rule of thumb, try to get rid of complicated terms first.

Let u = x² + 1. Then du = 2xdx.

Now we put u in the original problem and get

This won’t work. We still have x in it. But this is an easy task. We can add a constant.

It’s easy to work out an anti-derivative of u3, but notice that the original problem is all about x. Here we’ll change that x to u too. As a result the original problem becomes:

In conclusion:

Let’s try another example. This time we’ll just find an anti-derivative.

This time let u = x+1 won’t do. We need to make .

Thus x = u³ – 1. . Now we can get

Although it’ll become quite complicated, once we replace u in terms of x we get the answer.

(2) integration by parts

This is kind like product rule in reverse.

We’ll use this equation.

Check to see this equation is correct.

By sum rule we can get

Now let u = f(x), v = g(x).

Then du = f’(x)dx, dv = g’(x)dx.

Thus we can write the above equation to

The idea of integration by parts is that it lets you differentiate part of the integral at the price of anti-differentiating the other part of the integral.

The meaning of the above transformation is that we change one ∫ problem to another ∫ problem. The left side is the original problem, the right side is what we have to solve.

Let’s see an example.

Let u = x, then dv must equal to e^xdx. We’ve patched the left side.

Since u = x, we have . That is du = dx.

What about v? dv = e^xdx. We change it to more sensible look.

The problem now becomes what’s the anti-derivative of e^x? It’s also e^x. You can look up the reason on the web. Thus we have v = e^x. Now we just have to plug in the variables.

That’s the answer. You can actually differentiate the answer to see whether it works. Using product rule it’s plain enough.

And we can generalize this example a bit.

If we can change the original problem to something look like this:

We can let poly = u and e^xdx = dv. In this case we can always work out the answer.

(3) u + part

For some really hard problem, we need to combine u-substitution and integration by part.

Let  . Change the look to get .

Since , we have dx = 2udu. The original problem becomes

When you see e^x, you should always remember integration by part.

Let v = 2u. To patch we make dw = e^udu.

From the v part we have . Thus dv = 2du.

From the dw part we have . Thus w = e^u.

Now using the equation for integration by part we have

Finally we replace u with real value and get

2. Applications of integration

Here’re a few common usage for integration.

(1) Area

We’ve been calculating area the whole time. Now we’ll try something harder.

Question: what’s the area between ?

The first step is to build a model of the problem. Usually we’ll draw a graph and explicitly draw the Riemann rectangles on it. Here’s an example.

We can work out the crossing point is (1, 1) and (0, 0). What next?

To calculate the area between, we need to know each tiny rectangle’s width and height.

The width is dx, but what about the height?

We can pick a particular rectangle and see how the height is calculated. The upper bound is  the lower bound is x². Thus  is the height.

Now we can write the total area of rectangles in terms of integration.

The second step is to calculate the integration.

(2) volume

Integrations can help us calculate volumes.

Question: how to calculate the volume of a cone?

There’s a formula for this task. However, we want to use integration to get the task done, as a way of training out mind. Given a cone like this, what’s the volume?

The procedures are like area. We need to build a model first.

The key insight is that we can cut the cone into very tiny pieces. For each piece the height is dx and radius is r. Thus the volume is πr²dx.But we need to express r in terms of x. Now we put the cone into coordinate.

Since the upper line passes through (2, 1) and (0, 0), we can get its function .

This gives the answer for r. . Now we can do integration.