In this recitation, we covered problems 1,2,4,5,6,7,8, and 9 from “A Review of Integration.” I’ll discuss some of the important topics I wanted you to take away from these problems.

Fundamental Theorem of Calculus

A lot of the problems are related to the fundamental theorem of calculus. Remember that this says if \(f(x)\) is a continuous function on the interval \([a,b]\), then

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

where \(F'(x) = f(x)\). This is an extremely useful tool, both conceptually and when we want to get our hands dirty. Here are some example problems.

Problem: Is

\[\frac{d}{dt}\int_{t}^{t^2} e^{-x^2} dx\]

a function of \(t\), a constant, or neither?

Solution: In both recitations, I got the answer that this would be a function of \(t\) because of the bounds. For the most part, this is right, but let’s do a little more here. By the fundamental theorem of calculus, we can write

\[\int_t^{t^2} e^{-x^2} dx = F(t^2) - F(t),\]

where \(F'(x) = e^{-x^2}\). As we discussed, you’re going to be in trouble if you want to actually figure out what \(F(x)\) is. Luckily, we don’t need to for this problem. If we take the derivative of both sides, we have

\[\frac{d}{dt}\int_{t}^{t^2} e^{-x^2} dx = 2t F'(t^2) - F'(t).\]

Notice I had to use the chain rule here. Using the fact that \(F'(x) = e^{-x^2}\), we get

\[\frac{d}{dt}\int_{t}^{t^2} e^{-x^2} dx = 2t e^{-(t^2)^2} - e^{-t^2} = 2t e^{-t^4} - e^{-t^2}.\]

Notice the right hand side is a function of \(t\).

Problem: Evaluate

\[\int_0^2 xe^{-x^2}dx.\]

Solution: As we discussed in recitation, there are two ways to approach this problem. In both scenarios, we start with a \(u\)-sub. In this case, let’s let \(u = -x^2\), so \(du = - 2x dx.\) If we do this, then we can write

\[\int xe^{-x^2}dx = - \frac{1}{2}\int e^udu.\]

Notice how I didn’t write the bounds on the left hand side. This is me being careful – I don’t want to accidentally write the bounds in terms of \(x\) when I’m integrating with respect to \(u\). This is where the two paths diverge. One way of solving this problem is using the fundamental theorem of calculus. Using the right hand side, we can find the indefinite integral of \(xe^{-x^2}\):

\[- \frac{1}{2}\int e^udu = - \frac{1}{2} e^u + C = - \frac{1}{2} e^{-x^2} + C.\]

Notice how I didn’t say “the antiderivative of \(xe^{-x^2}\).” Remember that \(C\) is telling us there is a family of antiderivatives here, not just one. By choosing my favorite constant for \(C\), I get one antiderivative. The easiest constant to work with here is \(C=0\), so one antiderivative of \(xe^{-x^2}\) is \(- \frac{1}{2} e^{-x^2}\). By the fundamental theorem of calculus:

\[\int_0^2 xe^{-x^2}dx = -\frac{1}{2} e^{-x^2} \bigg|_{x=0}^2 = - \frac{1}{2} \left[e^{-2^2} - e^{-0^2} \right].\]

The other way of solving this problem involves changing the bounds. Remember that \(du\) means my bounds should be in terms of \(u\). If \(x = 0\), then we have \(u = -0^2 = 0\), and if \(x=2\) then \(u = -2^2 = -4\). So I can rewrite my integral as

\[\int_0^2 xe^{-x^2}dx = - \frac{1}{2}\int_0^{-4} e^udu.\]

Now I use the fundamental theorem of calculus here, but with just my \(u\)-variable instead. I get

\[- \frac{1}{2}\int_0^{-4} e^udu = - \frac{1}{2} e^u \bigg|_{u=0}^{-4} = - \frac{1}{2} \left[e^{-4} - e^0 \right].\]

Either way, I get the right answer!

Problem: Can we use the fundamental theorem of calculus to solve

\[\int_{-1}^1 \sqrt{x}dx?\]

Solution: This one is easier than the recitation problem: \(\sqrt{x}\) isn’t even defined when \(x < 0\), so I cannot use the fundamental theorem of calculus.

U-sub

The only other note I want to make about \(u\)-sub is that we need to be careful of when we can apply it. Here are a few example problems.

Problem: Solve the following integrals:

1) \(\int \frac{dx}{x^2 + 4},\)

2) \(\int \frac{xdx}{x^2 + 4},\)

3) \(\int \frac{x^2dx}{x^2 + 4}.\)

Solution: This one is a little tricky. Hopefully we remember that

\[\frac{d}{dx} \arctan(x) = \frac{1}{x^2 + 1},\]

but notice we have a \(4\) instead of a \(1\) in our denominators. What do we do?

Well, let’s try factoring out a \(4\). By factoring out, I mean dividing everything by \(4\), and then multiplying all of the terms by \(4\):

\[x^2 + 4 = 4 \left( \frac{x^2}{4} + 1 \right).\]

Even with this, the denominator still isn’t quite right. However, notice there’s more algebra I can do here. Using the fact that \(2^2 = 4\), we can write

\[x^2 + 4 = 4 \left( \left(\frac{x}{2}\right)^2 + 1 \right).\]

This is more promising, but I wanted just \(x^2\) and not \((x/2)^2\). What we can do to transform it into this nicer format is to create a dummy variable. Let \(u = x/2\), then

\[x^2 + 4 = 4 \left( u^2 + 1 \right).\]

Going back to the first integral, we can now write

\[\int \frac{dx}{x^2 + 4} = \frac{1}{4} \int \frac{dx}{u^2 + 1}.\]

This is inconvenient because I am integrating with respect to \(x\) and I have a \(u\). Notice that

\[du = \frac{dx}{2},\]

so

\[\int \frac{dx}{x^2 + 4} = \frac{1}{2} \int \frac{du}{u^2 + 1} = \frac{1}{2} \arctan(u) + C.\]

Finally, I need my answer in terms of \(x\), so

\[\int \frac{dx}{x^2 + 4} =\frac{1}{2} \arctan(u) + C = \frac{1}{2} \arctan(x/2) + C.\]

This solves the first integral.

For the second integral, we don’t need to get as fancy. We can just do a \(u\)-substitution here:

\(u = x^2 + 4, \ \ du = 2xdx\),

so

\[\int \frac{xdx}{x^2 + 4} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln(\mid u \mid) + C = \frac{1}{2} \ln(x^2 + 4) + C.\]

For the last integral, a \(u\)-substitution is not helpful. Instead, we can do a common trick in math: adding \(0\). This seems silly, but to get even more silly we can write \(0 = 4 - 4\). So

\[\int \frac{x^2dx}{x^2 + 4} = \int \frac{x^2 + 4 - 4}{x^2 + 4}dx = \int \frac{x^2+ 4}{x^2 + 4}dx - \int \frac{dx}{x^2 + 4}.\]

Simplifying this gives us

\[\int \frac{x^2dx}{x^2 + 4} = x - \frac{1}{2} \arctan(x/2) + C.\]