The first problem on today’s quiz is a definite integral. In order to find the exact area, we’ll rely on the Fundamental Theorem of Calculus — finding first an antiderivative of \(f(x)=\sqrt[3]{x}\), and then evaluating the antiderivative at \(x=27\) and \(x=1\), then subtracting the two results.
First, we consider \(\sqrt[3]{x}\) as \(x^\frac13\) in order to apply the “anti-power rule”, finding \(F(x)=\frac34x^\frac43\). Then \(F(27)\) simplifies to \(\frac{243}{4}\), and \(F(1)\) simplifies to \(\frac34\). The exact area, then, is \(F(27)-F(1)=60\).
In the second problem, we again see a definite integral, this time with \(f(x)=\frac{6}{\sqrt[3]{x}}=6x^{-\frac13}\). Again, applying the power rule for antiderivatives, \(F(x)=\frac{6x^\frac23}{\frac23}=9x^\frac23\). With \(F(27)=81\) and \(F(1)=9\), we get an exact area of \(72\).
For our third problem, extra care must be taken to manipulate \(f(x)\) into a form that we can again use the power rule for antiderivatives. First, we expand the numerator: \((x+2)^2 = x^2+4x+4\), then divide by \(x^2\) to end up with \(f(x)=1+\frac{4}{x}+\frac{4}{x^2}\). We then look to find an antiderivative for each of these three terms:
- \(x\) as the antiderivative of \(1\),
- \(4\ln|x|\) as the antiderivative of \(\frac{4}{x}\),
- and \(-4x^{-1}\) as the antiderivative of \(4x^{-2}\).
With \(F(x)=x+4\ln|x|+\frac{-4}{x}\), all that remains is to subtract \(F(2)-F(1)=(2+4\ln 2-2)-(1+4\ln 1-4)=3+4\ln 2\)
The last two problems focus on the antiderivative(s) of \(f(x)=8\sin(x)+4e^x\).
First, we find the general form of the antiderivative: \(F(x)=-8\cos(x)+4e^x+C\)
Then, with the additional requirement that \(F(0)=5\), we can determine the specific antiderivative that satisfies this requirement:
- \(F(0)=-8\cos(0)+4e^0+C\)
- \(5=-8+4+C\)
This means that \(C=9\) will result in the exact antiderivative that passes through the required point: \(0,5\). \(F(x)=-8\cos(x)+4e^x+9\)
Antiderivatives are unavoidably tied to derivatives. Because of this connection, it is crucial to understand derivative rules as we proceed with similar rules for antiderivatives. We start with the antiderivative version of the chain rule — so a refresher on the chain rule is in order.
For example, when we look to take the derivative of \(\sin(2x)\), we first observe that this is a composition of two functions: \(f(x)=\sin(x)\) and \(g(x)=2x\), so that \(f(g(x))=\sin(2x)\). Chain rule applies to composition of functions such as this. So the derivative of \(\sin(2x)\) is \(f'(g(x))g'(x)=\cos(g(x))\cdot 2=2\cos(2x)\).
In general, for antiderivatives, when we see composition of functions, we will use a strategy known as “u-substitution.” We will introduce \(u\) as a new variable to take the place of the \(g(x)\) inside the composed functions. As part of this substitution, we are also required to convert the differential \(dx\) over to \(du\).
The relationship between \(dx\) and \(du\) is formed from the relationship between \(u\) and \(x\): \(u=g(x)\) so \(du=g'(x)\,dx\). Let’s see what that looks like in practice.
When attempting to find the antiderivative: \(\int 2\cos(2x)\,dx\), we identify \(u=2x\) as the inside function in the composition. With \(\color{green}{u=2x}\), then \(\color{blue}{du=2\,dx}\).
Now we can observe that \(\color{blue}2\,dx\) is already present in \(\int \textcolor{blue}{2}\cos(\textcolor{green}{2x})\,\textcolor{blue}{dx}\), meaning that after substituting we have \[\int\cos(\textcolor{green}{u})\,\textcolor{blue}{du}=\sin(\textcolor{green}{u})+C\]
This is the antiderivative of the cosine function, which we know to be sine. So that is \(\sin(u)+C\), but we know that since we started with a function of \(x\) we must also end up with an antiderivative that is a function of \(x\). That means we must return \(u\) back to being \(2x\). Our final answer for the antiderivative is \(\sin(2x)+C\).
In our next example, we aim to find the antiderivative: \(\int\sqrt[3]{x+1}\,dx\). There are four common places where we usually look for substitution targets:
- what’s in parenthesis
- what’s in a denominator
- what’s under a radical
- what’s in an exponent
Here, we have something more complicated (not by much) than just \(x\) under a radical. That means we will attempt to substitute this radicand with \(u=x+1\), which has differential \(du=1\,dx\). This is the easiest situation possible, as it means we may replace \(dx\) with \(du\). So after the substitution, we have \(\int\sqrt[3]{u}\,du\), which we already found (check back to the first quiz problem). Then the antiderivative is \(\frac34 u^\frac43+C\), which must be turned back into a function of \(x\): \[F(x)=\frac34(x+1)^\frac43+C\]
Now, just because we start a substitution doesn’t mean that it will be successful (that is to say, that we’ll be able to complete the antiderivative after substituting). I provide this as an example of an attempted substitution that fails. It’s important to know when our plan has gone off the rails.
In this example, we have the antiderivative \(\int e^{(x^2)}\,dx\). Following our previous advice, here we see an exponent that is more complicated than just \(x\). That means our obvious target for substitution would be the exponent: \(u=x^2\).
If \(u=x^2\), then our differential is \(du=2x\,dx\). We don’t see \(2x\,dx\) in our integral, but we can force the issue by replacing \(dx=\frac{1}{2x}\,du\) (solving for \(dx\)). After substituting, we have \(\int e^u\frac{1}{2x}\,du\) — but our substitution is still incomplete because we still have an \(x\) in our integral. Now this lone \(x\) is (almost) impossible for us to resolve with substitution. We can again attempt to force the issue by isolating \(x\) from our \(u=x^2\) substitution: \(x=\pm\sqrt{u}\).
I wrote out the result, but we don’t even know whether we should be replacing this remaining \(x\) with \(\sqrt{u}\) or \(-\sqrt{u}\). The point is that even if we did, this integral is now a complete dumpster fire.
In our last example, we look at \(\int\frac{x+1}{x^2+2x+10}\,dx\). Again, pulling from our list of places where we can find substitutions, we observe a denominator \(u=x^2+2x+10\) with a very nice differential (for what we’re working with): \(du=2x+2\,dx\). Why is this nice for our integral?
Look at our numerator, we have \(x+1\) with a \(dx\) factor. When doubled, this is exactly the differential: \(du=2(x+1)\,dx\), which can be rewritten as \(\frac12\,du=x+1\,dx\). So our substituted integral becomes \[\int\frac{1}{u}\cdot\frac12\,du=\frac12\int\frac{1}{u}\,du=\frac12\ln|u|+C\]
Converting this back to being a function of \(x\), we get \[F(x)=\frac12\ln|x^2+2x+10|+C\]
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