6.4: Fundamental Theorem of Calculus
In the previous two sections, we saw that the area under a curve can be found using more and more rectangles. However, this process can be tedious and not very enlightening. There is a powerful theorem that allows us to compute area under the curve quickly in many cases.
Given a function \(f(x)\) where \(F(x)\) is an anti-derivative of \(f(x)\), we have
\(\int_a^b f(x)dx = F(b) - F(a)\)
This is a really remarkable theorem. At first blush, finding the area under a curve and finding the slope of a tangent line have nothing in common. What this theorem is saying is that these are intimately tied, and in fact that are exactly inverse operations. That is, one undoes the other. Moreover, this gives an exact answer to integral problems, something that eluded us in the previous sections.
For example, let’s say we wanted to solve the following:
To use the fundamental theorem, we need an anti-derivative of \(f(x) = x^2\). The anti-derivative in this case, by the inverse power rule, is \(F(b) - F(a)\), which is
\[\begin{align*} \int_2^5 x^2dx & = \frac{5^3}{3} - \frac{2^3}{3} \\ & = \frac{125}{3} - \frac{8}{3} \\ & = \frac{117}{3} \\ & = \boxed{39} \end{align*}\]
The area under the curve is \(39\). Note that \(F(b) - F(a)\) is sometimes denoted \(F(x) \Big|_a^b.\) We’ll see that in the following examples.
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Find \(\int_{-2}^1 3x^2dx\).
To solve this, we find the anti-derivative of \(3x^2\), and then plug in the end points and subtract the result. First note that \(\int 3x^2dx = x^3 + C\).
\[\begin{align*} \int_{-2}^1 3x^2dx & = x^3 + C \Big|_{-2}^1 \\ & = (1^3 + C) - ((-2)^3 + C) \\ & = 1 + C - (-8 + C) \\ & = 1 + C + 8 - C \\ & = \boxed{9} \end{align*}\]
Notice how the constants of integration cancel? For definite integral problems, we can essentially ignore the constant of integration for this reason.
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Find \(\int_0^6 \frac{1}{2}x^2 + \frac{1}{3} xdx\).
We integrate and then evaluate again.
\[\begin{align*} \int_0^6 \frac{1}{2}x^2 + \frac{1}{3} xdx & = \frac{1}{2} \int_0^6 x^2 + \frac{1}{3} \int_0^6 xdx \\ & = \left( \frac{1}{2} \frac{x^3}{3} + \frac{1}{3} \frac{x^2}{2} \right) \Big|_0^6 \\ & = \left( \frac{x^3}{6} + \frac{x^2}{6} \right) \Big|_0^6 \\ & = \frac{x^3 + x^2}{6} \Big|_0^6 \\ & = \left( \frac{6^3 + 6^2}{6} \right) - \left( \frac{0^3 + 0^2}{6} \right) \\ & = \frac{252}{6} - \frac{0}{6} \\ & = \boxed{42}. \end{align*}\]
Why does the fundamental theorem of calculus work? As we have seen earlier, it is sometimes easiest to see in terms of position versus velocity. Let \(F(x)\) be the position function of a car, let \(f(x)\) be the velocity function of the car, and let \(x\) be the time in hours. For simplicity, let’s say \(f(x)\) is constant from \(x = 2\) to \(x = 6\):
So if this is the velocity function, what is happening to the position function? As we’ve seen, we just need to multiply : its the \(10 \frac{\text{miles}}{\text{hour}}\) times the four hours, gives a change of 40 miles. That means \(F(x)\) looks something like this:
Note it need not start at \(5\) miles, but that is one possibility.
Now let’s look at the fundamental theorem again:
Given a function \(f(x)\) where \(F(x)\) is an anti-derivative of \(f(x)\), we have
\(\int_a^b f(x)dx = F(b) - F(a)\)
Can you see why the fundamental theorem worked out in this case? We see see \(F(b) - F(a)\) is just the change of position between hour \(2\) and hour \(6\), which is \(45-5\) or \(40\) miles. What is \(\int_a^b f(x)dx\)? Well, that’s the area under the velocity curve. How do you find area? That’s right, it’s multiplication ! In particular, it is \(10 \frac{\text{miles}}{\text{hour}}\) times the four hours, again giving a change of \(40\) miles.
You can see how both area under the curve and antiderivatives come down to the same basic calculation. That’s why the fundamental theorem of calculus can claim they are the same thing.
If the velocity function is more complicated, this still works. We can think of a more complicated function as a combination of these constant functions.