The second fundamental theorem of calculus, sometimes called the second FTC, is a formula that allows for the evaluation of definite integrals. In this article, we will explore what the second half of the fundamental theorem of calculus states, how it functions in practice, and how it is connected to the first fundamental theorem of calculus.

It is possible that your teacher or textbook call this theorem “The first fundamental theorem of calculus”, as there is no consensus – we have addressed this in detail in another article, linked here: Fundamental Theorem of Calculus: which is first and which is second?. For information on the other half of the fundamental theorem of calculus, check out First Fundamental Theorem of Calculus.

#### What is the Second Fundamental Theorem of Calculus?

The second fundamental theorem of calculus states the following:

Given a function \(f(x)\) continuous on the interval \([a,b]\) and another function \(F(x)\) such that \(F'(x) = f(x)\), then

\[\int_a^b f(x) \,dx = F(b) – F(a)\]

This means that if we are either given the antiderivative of \(f(x)\), or we can calculate that antiderivative, then we can find the definite integral, or area under the curve, of \(f(x)\).

#### Historical Background

Because both parts of the fundamental theorem of calculus were formulated together, this theorem shares the same history as the first fundamental theorem of calculus. They were independently formulated by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century. Both scientists are credited for the discovery and reached the conclusion through wildly different methods. Newton approached calculus through his method of fluxions, while Leibniz developed a systematic notation that is still in use today. While Newton discovered it first, Leibniz was the first to publish the findings, but they are both credited with the discovery, read more about it in Who Discovered Calculus?

#### Example

Let’s say we want to evaluate the following definite integral:

\[\int_1^4 (x^2 – 1) \,dx\]

To do this, we first need to find the antiderivative of \(x^2 – 1\). We will use the Power Rule to determine the said antiderivative. The antiderivative of \(x^2\) is \(\frac{x^3}{3}\), and the integral of \(-1\) is \(-x\). We can check this by differentiating \(\frac{x^3}{3}\) and \(-x\). We thus find that an antiderivative of \(x^2 – 1\) is \(\frac{x^3}{3} – x\).

It is important to note that \(x^2 -1\) has many antiderivatives. For example, \(\frac{x^3}{3} – x + 8\), \(\frac{x^3}{3} – x-\pi\), \(\frac{x^3}{3} – x + 2\sqrt{2}\) are all antiderivatives of \(x^2 – 1\). That is why when we calculate the indefinite integrals, we always add \(+C\) at the end. However, we do not need to concern ourselves with this, since we are calculating the definite integral.

According to the second fundamental theorem of calculus,

\[\int_1^4 (x^2 – 1) \,dx = F(4) – F(1),\]

where \(F(x) = \frac{x^3}{3} – x\). Thus, to evaluate the integral, we need to evaluate \(F(x)\) at \(x = 1\) and \(x = 4\). Plugging in those values, we find that \(F(4) = \frac{52}{3}\) and \(F(1) = \frac{-2}{3}\). The definite integral is equal to:

\[F(4) – F(1) = \frac{52}{3} – \frac{-2}{3} = \boxed{18}\]

#### A Common Misconception

One misconception that may be encountered is forgetting to check if \(f(x)\) is continuous over \([a,b]\). As a reminder, \(f(x)\) must be continuous between the bounds of the definite integral in order for the second FTC to be applicable. Let’s say we want to find the area under \(y = f(x)\) on the interval \([0,6]\):

Since \(f(x)\) is not continuous on the interval, we cannot find the antiderivative of \(f(x)\), and thus we cannot find the area under \(f(x)\). If we wanted to find the area under this curve, we would instead have to realize that \(f(x)\) is a piecewise function split the interval of integration:

\[ f(x) = \begin{cases} 3 & \text{if } x < 2 \\ 7 – x & \text{if } x \geq 2 \end{cases} \]

\[\int_0^6 f(x) \,dx = \int_0^2 f(x) \,dx + \int_2^6 f(x) \,dx = \]

\[\int_0^2 3\, dx + \int_2^6 (7-x)\,dx\]

Since \(f(x)\) is continuous on the intervals \([0,2)\) and \([2,6]\), we can evaluate the integrals on each interval and then add them to find the area under \(y=f(x)\).

#### Conclusion

The second fundamental theorem of calculus establishes a connection between definite and indefinite integrals. While indefinite integrals, also known as antiderivatives, are functions, definite integrals, which are equal to the area under a graph of the function, are values. It should be surprising that these two concepts are related. In this article, we discussed the second part of the theorem. Definite integrals can also be viewed as accumulation functions, which we did not discuss in this article. It is important to keep in mind that different areas of calculus may be related in some very surprising ways and the fundamental theorem of calculus is an example of such a relationship.