Evaluating Definite Integrals with the Fundamental Theorem of Calculus, Part 2

Hello!

In our last lesson, we formally defined the definite integral as the limit of a Riemann sum. We established that $\int_{a}^{b} f (x) d x$ represents the exact signed area under the curve of $f (x)$ from $x = a$ to $x = b$ . We also acknowledged that calculating this limit directly is incredibly tedious and impractical.

Today, you'll learn the remarkably elegant and powerful method for evaluating definite integrals. This lesson introduces the Fundamental Theorem of Calculus, Part 2, which connects the concept of area (integration) directly back to the process of anti-differentiation that we've already studied. This theorem is one of the most important results in all of mathematics.

Your learning outcome is to evaluate definite integrals using the Fundamental Theorem of Calculus, Part 2.

1. The Bridge Between Antiderivatives and Area

The core idea of the Fundamental Theorem is astonishingly simple: the definite integral of a function over an interval can be found by finding an antiderivative of that function and simply evaluating it at the endpoints of the interval.

Let's see the formal statement of this theorem.

The Fundamental Theorem of Calculus

This resource from technologyuk.net provides a clear and direct statement of the Second Fundamental Theorem of Calculus.

Please read the section titled 'The second fundamental theorem of calculus'. Focus on understanding the core formula presented.

As the resource states, if $f (x)$ is a continuous function on the interval $[a, b]$ and $F (x)$ is any antiderivative of $f (x)$ (meaning $F^{'} (x) = f (x)$ ), then:

\int_{a}^{b} f (x) d x = F (b) - F (a)

This is a revolutionary result! Instead of summing an infinite number of infinitesimal rectangles, we just need to:

Find an antiderivative, $F (x)$ .
Evaluate it at the upper limit, $b$ .
Evaluate it at the lower limit, $a$ .
Subtract the results.

A common shorthand notation for $F (b) - F (a)$ that you will see frequently is:

[F (x)]_{a}^{b} or F (x)_{a}^{b}

What about the Constant of Integration, C?

When we find an indefinite integral, we always add a constant $C$ . You might be wondering why it's missing here. The following reading explains this crucial point.

The Fundamental Theorem of Calculus

Let's clarify why the constant of integration, '+ C', is not needed for definite integrals.

Read the short section beginning 'The definite integral is a geometric concept...'. The key takeaway is seeing how the constant 'C' always cancels out when you calculate F(b) - F(a).

Because the constant $C$ always cancels out ( $(F (b) + C) - (F (a) + C) = F (b) - F (a)$ ), we can simply use the simplest antiderivative—the one where $C = 0$ .

2. A Step-by-Step Example

Let's walk through the entire process with a concrete example. The image below provides a clear, step-by-step solution.

Evaluating a Definite Integral using the Fundamental Theorem of Calculus, Part 2 — This image shows the full process for evaluating $\int_{-1}^{2} (2x+1)^2 dx$. Let's break down the steps shown.

Here is a textual breakdown of the process in the image:

Prepare the Integrand: The function to be integrated (the integrand) is $(2 x + 1)^{2}$ . To make it easy to find an antiderivative, we first perform the algebra to expand it: $(2 x + 1)^{2} = 4 x^{2} + 4 x + 1$ . Our integral is now $\int_{- 1}^{2} (4 x^{2} + 4 x + 1) d x$ .
Find an Antiderivative: Using the power rule on each term of the polynomial, we find an antiderivative, $F (x)$ .
$F (x) = \frac{4 x ^{3}}{3} + \frac{4 x ^{2}}{2} + x = \frac{4}{3} x^{3} + 2 x^{2} + x$
Apply the Theorem (Evaluate at Limits): Now we calculate $F (2) - F (- 1)$ .
- First, evaluate at the upper limit, $b = 2$ :
  $F (2) = \frac{4}{3} (2)^{3} + 2 (2)^{2} + 2 = \frac{4}{3} (8) + 2 (4) + 2 = \frac{32}{3} + 8 + 2 = \frac{32}{3} + 10 = \frac{62}{3}$
- Next, evaluate at the lower limit, $a = - 1$ :
  $F (- 1) = \frac{4}{3} (- 1)^{3} + 2 (- 1)^{2} + (- 1) = - \frac{4}{3} + 2 - 1 = - \frac{4}{3} + 1 = - \frac{1}{3}$
Subtract:
$F (2) - F (- 1) = \frac{62}{3} - (- \frac{1}{3}) = \frac{62}{3} + \frac{1}{3} = \frac{63}{3} = 21$

The value of the definite integral is 21. This number represents the net area under the curve of $y = (2 x + 1)^{2}$ between $x = - 1$ and $x = 2$ .

3. Practice with Various Functions

The power of the Fundamental Theorem lies in its broad applicability. As long as you can find an antiderivative, you can evaluate the definite integral. The following video provides excellent worked examples covering various types of functions.

Fundamental Thereom of Calculus Explained - Part 1 & 2 Examples - Definite Integral

Now that you've seen the process, let's watch some more worked examples to solidify your understanding. This video by The Organic Chemistry Tutor demonstrates how to apply the theorem to functions you've encountered before.

Watch from 10:42 to 21:56. The presenter works through several examples, from simple polynomials to functions with fractional and negative indices (exponents). Pay close attention to the consistent process: find the antiderivative, plug in the upper limit, plug in the lower limit, and subtract.

Test your understanding!

Evaluate the definite integral $\int_{1}^{4} (3 x + \frac{1}{x ^{2}}) d x$ .

Hint: First, rewrite the function using indices (exponents) before finding the antiderivative.

Show answer

First, rewrite the integrand with indices: $f (x) = 3 x^{1/2} + x^{- 2}$ .

Next, find the antiderivative $F (x)$ :
$F (x) = 3 \cdot \frac{x ^{3/2}}{3/2} + \frac{x ^{- 1}}{- 1} = 2 x^{3/2} - x^{- 1} = 2 x^{3/2} - \frac{1}{x}$
Now, evaluate using the notation $[F (x)]_{1}^{4}$ :
$[2 x^{3/2} - \frac{1}{x}]_{1}^{4} = (2 (4)^{3/2} - \frac{1}{4}) - (2 (1)^{3/2} - \frac{1}{1})$
Calculate each part:
$(2 (4)^{3} - \frac{1}{4}) - (2 - 1)$
$(2 (2)^{3} - \frac{1}{4}) - (1)$
$(16 - \frac{1}{4}) - 1 = \frac{63}{4} - 1 = \frac{59}{4}$
The final answer is $\frac{59}{4}$ or $14.75$ .

The theorem also applies seamlessly to other function families, such as trigonometric functions.

Fundamental Thereom of Calculus Explained - Part 1 & 2 Examples - Definite Integral

Let's continue with the same video to see how the theorem is applied to trigonometric functions.

Watch from 21:56 to 25:05. This segment covers examples with cos(θ) and sec²(θ). Notice that the process is identical; success simply depends on recalling the correct antiderivatives.

4. Properties of Definite Integrals

Just like with derivatives and indefinite integrals, definite integrals have properties that can help simplify calculations. The reading below, from a module by the University of Reading, lists these properties.

PPLATO | FLAP | MATH 5.2: Basic integration

This resource lists several useful properties of definite integrals. Understanding these can sometimes save you a lot of work.

Please read the section '4.1 Properties of definite integrals'. Pay attention to properties (19), (20), and (21), as they are particularly useful in algebraic manipulation.

Here are three of the most important properties from that list:

Reversing Limits: Swapping the limits of integration negates the value of the integral.
$\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x$
Additivity: You can split an integral into two parts. This is useful for piecewise functions or when analysing different segments of an area.
$\int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x$
Linearity: You can pull out constant multiples and split sums/differences, just as with indefinite integrals.
$\int_{a}^{b} [k \cdot f (x) \pm l \cdot g (x)] d x = k \int_{a}^{b} f (x) d x \pm l \int_{a}^{b} g (x) d x$

This linearity property is what allowed us to find the antiderivative term-by-term in our examples.

Conclusion

Today we've unlocked the primary tool for computing definite integrals. What was once a daunting infinite-sum problem is now a straightforward process of anti-differentiation and substitution.

Key Takeaways:

The Fundamental Theorem of Calculus, Part 2 states that $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , where $F$ is any antiderivative of $f$ .
The process is to find the antiderivative, evaluate it at the upper and lower limits, and subtract.
The constant of integration $C$ is not needed for definite integrals because it always cancels out.
The theorem applies to all continuous functions for which an antiderivative can be found, including polynomial, power, and trigonometric functions.

In our next lesson, we will explore the other half of this theorem: the Fundamental Theorem of Calculus, Part 1. It explores the deep relationship between differentiation and integration from another angle, defining the integral as a function. This will solidify your understanding of why these two seemingly different branches of calculus are so intimately connected.