On today’s quiz, a graph of \(f(x)\) was given, and everyone was asked to evaluate five definite integrals. In our first case, we are looking for the area on the interval \([0,3]\). This region is comprised of a right triangle from \([0,2]\), and a rectangle on \([2,3]\). The areas are each \(2\) square units, for a total area of 4.
Then we are asked for the area on the interval \([0,5]\), and we know that the area on \([0,3]\) is \(4\); so we just need to determine the area on \([3,5]\). Now, our function crosses the \(x\)-axis on this interval, and we have two ways to think of this kind of situation. The most obvious is that we have two right triangles, one positive (on the left) and one negative (on the right). What is hard about this perspective is that we must determine the location of the \(x\)-intercept (in order to find the width of each of the two triangles), and that’s not so easy to do. (We talked it through, and the positive area is \(\frac43\) and the negative is \(-\frac13\) for a total area of \(1\).) Instead, we can view this area similar to other trapezoids, combining a right triangle with a rectangle — though in this configuration, we subtract the rectangle from the right triangle (to represent dropping a positive triangle below the \(x\)-axis). The triangle has area \(3\), and it is dropped down \(1\) unit across a width of \(2\), so we subtract a rectangle of area \(2\); again for a total of \(1\). Combining this \([3,5]\) area with the \([0,3]\) area, we end up with \(\displaystyle\int_0^5 f(x)\,dx=5\)
The next definite integral covers the interval \([7,9]\), a negative rectangle with a negative triangle: \(-6 + -1 = -7\)
The fourth definite integral covers \([5,9]\), only needing to find \([5,7]\) (since we know the area over \([7,9]\)). On \([5,7]\), we again have a negative rectangle with a negative triangle: \(-2 + -3 = -5\). Combining the two intervals, we have \(\displaystyle\int_5^9 f(x)\,dx=-12\)
Finally, the area over the entire (graphed) interval: \([0,9]\), which we can get by combining the areas over \([0,5]\) and \([5,9]\): \(5+ -12 = -7\)
We started today’s lesson by quickly running through the manner in which we would use Riemann Sums to approximate the area under the curve \(f(x)=x^p\) (where \(p\) is a positive integer) over the interval \([0,a]\) with \(n\) rectangles.
- First we compute \(\Delta x = \frac{a-0}{n} = \frac{a}{n}\)
- Then the \(x\)-values: \(x_i = \frac{a}{n}\cdot i\)
- Next, the \(y\)-values: \(y_i = f(x_i) = \frac{a^p}{n^p}\cdot i^p\)
- Finally, the areas: \(A_i = y_i \cdot \Delta x = \frac{a^{p+1}}{n^{p+1}}i^p\)
We get the total area by summing the \(A_i\)s: \(\displaystyle\sum^n_{i=1}A_i = \frac{a^{p+1}}{n^{p+1}}\sum^n_{i=1}i^p\), noting that the common factor of \(\frac{a^{p+1}}{n^{p+1}}\) can be multiplied after summing the \(i^p\).
Before we continue, note that we want a closed form for the \(\sum i^p\) so that we can compute a limit as \(n\to\infty\).
Now, we didn’t go through how we might find a closed form for \(\sum i^p\), but we can find one for any \(p\) that is a positive integer. Moreover, the closed form will always be a polynomial with degree \(p+1\) and leading coefficient \(\frac{1}{p+1}\). So our total area is: \[A_n=\frac{a^{p+1}}{n^{p+1}}\sum^n_{i=1}i^p=\frac{a^{p+1}(n^{p+1}+\ldots(\text{lower-degree terms}))}{n^{p+1}(p+1)}\]
This \(A_n\) has the same degree (of \(n\)) in the numerator and denominator, so there will be a horizontal asymptote, and hence a limit as \(n\to\infty\).
So the exact area under \(f(x)=x^p\) on the interval \([0,a]\) is \(\lim A_n=\frac{a^{p+1}}{p+1}\). This result depends on \(p\) (the exponent of \(x\) in \(f(x)=x^p\)), and more importantly, on \(a\), the endpoint of our interval \([0,a]\).
Thinking of this area as a function of \(a\), we have:
- \(\displaystyle A(a)=\int_0^ax^p\,dx=\frac{a^{p+1}}{p+1}\)
- Exchanging the \(a\) for \(t\) (\(t\) is more clearly a variable): \(\displaystyle A(t)=\int_0^tx^p\,dx=\frac{t^{p+1}}{p+1}\)
- And then if we exchange the roles of \(x\) and \(t\): \(\displaystyle A(x)=\int_0^xt^p\,dt=\frac{x^{p+1}}{p+1}\)
This is just a little bit of rearranging of our representation — there’s no actual change to the meaning.
Here we work through the derivative of \(A(x)\), finding that \[\frac{d}{dx}\left[\frac{x^{p+1}}{p+1}\right]=x^p=f(x)\]
We conclude then, that the relationship between the area function \(A(x)\) and the original function \(f(x)\) is:
- \(f(x)\) is the derivative of \(A(x)\)
- \(A(x)\) is the antiderivative of \(f(x)\)
Now we look at the visual representation of our area function where \(f(x)\) could be any continuous function: \(A(x)=\int^x_0 f(x)\,dx\) and think about the derivative, \(A'(x)\). Using the limit definition, we’ll need to consider \(A(x+h)-A(x)\), which ends up being the area under \(f(x)\) on the interval \([x,x+h]\). Then in the limit definition of derivative for \(A(x)\), this area is divided by \(h\) (which is the width of the area), leaving us with just the height of the function (somewhere between \(f(x)\) and \(f(x+h)\), both of which converge to \(f(x)\) as \(h\to 0\)).
So we conclude that \(\displaystyle\frac{d}{dx}\left[\int_0^x f(t)\,dt\right]=f(x)\)
We call this result the “Fundamental Theorem of Calculus”:
- When we want to calculate \(\displaystyle\int_a^bf(x)\,dx\),
- IF we find a function \(F(x)\) such that \(F'(x)=f(x)\), (in other words, \(F(x)\) is an anti-derivative of \(f(x)\))
- THEN the definite integral can be calculated: \(\displaystyle\int_a^bf(x)\,dx=F(b)-F(a)\)
So it is going to be VERY important to find antiderivatives. In general, we will use the “indefinite” integral (no bounds) to represent the action of finding antiderivatives: \(\int f(x)\,dx = F(x) + C\).
Here, we note that for any function \(f(x)\), there is an infinite “family” of antiderivatives: \(F(x)+C\), since \(\frac{d}{dx}\left[F(x)+C\right]=F'(x)+0=f(x)\). This tells us that all antiderivatives in this infinite “family” have the same derivative.
So this “+C” does not affect the derivative being \(f(x)\). And this should make sense, as “+C” on a function has the effect of shifting the graph up or down identically across all \(x\)-values. This means that the shape of the curve (specifically, the values of the derivative — the instantaneous rate of change) is not affected.
Think carefully, does the slope of the line tangent to \(f(x)\ at \(x=0\) change if \(f(x)\) is shifted vertically (either up or down)? Would the same be true at other \(x\)-values? Hopefully, you are able to conclude (correctly) that vertical shifts do not affect the slope of the tangent line (which is the value of the derivative).
So, as we saw from our first example, \(f(x)=x^p\), we get the “anti” power rule for antiderivatives: \(\int x^p\,dx=\frac{1}{p+1}\cdot x^{p+1} + C\)
It is possible, in some situations, that we will know some specific point that our antiderivative must pass through — such as \((a,b)\). This means that when \(x=a\), we want the specific antiderivative that has \(y\)-value \(=b\). So, then \(b=F(a)+C\), where \(a\) and \(b\) are known values, leaving us with only C unknown. We can then find the value of \(C=b-F(a)\).
To wrap things up, we note that antiderivatives have the same linear properties as derivatives do:
- \(\frac{d}{dx}\left[k\cdot f(x)\right]=k\cdot\frac{d}{dx}\left[f(x)\right]\) so also does \(\int k\cdot f(x)\,dx = k\cdot\int f(x)\,dx\).
- \(\frac{d}{dx}\left[f(x)+g(x)\right]=\frac{d}{dx}\left[f(x)\right]+\frac{d}{dx}\left[g(x)\right]\) so also does \(\int \left(f(x)+g(x)\right)\,dx =\int f(x)\,dx + \int g(x)\,dx\).
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