#6.7 The Fundamental Theorem of Calculus and Definite Integrals

Welcome to one of the most important theorems in all of calculus! In this key topic, we’ll explain what the two parts of the fundamental theorem of calculus (FTOC) are as well as explain the relationship between a definite integral and its antiderivative!

#1️⃣ Fundamental Theorem of Calculus Part 1

The first part of the FTOC states the relationship between antiderivatives and definite integrals. Now what does this mean? 🤔 Let’s write this out.

$g(x)=\int_{a}^{x} f(t) dt$

$g'(x)=f(x)$

What the equation above represents is that if you take the derivative of an integral, you will be left with the inside function. This happens because differentiation and integration cancel each other out, leaving you with the inside function.

However, you need to be cautious! If the upper bound is not just $x$ , then you will have to substitute whatever the upper bound is for $t$ .

#✏️ Example 1 - Normal FTOC

Find $g'(x)$ .

$g(x)=\int_{32}^{x} 5t^4 dt$

Using the FTOC, you can simplify the right side by taking the derivative of both sides. Therefore, $g'(x)=5x^4$ .

#✏️ Example 2 - Upper Bound Change

Find $g'(x)$ .

$g(x)=\int_{32}^{2x} 5t^4 dt$

Note that in this example, the upper bound is now $2x$ . The only change that you will do to find $g'(x)$ will be to substitute $2x$ for $t$ , instead of $x$ . Therefore $g'(x)=5(2x)^4=80x^4$ .

Untitled

A visual representation of the first part of the FTOC

Image Courtesy of Math24

📌 We actually already discussed this part of the fundamental theorem of calculus in key topic 6.4. For more information and examples, check it out here.

#2️⃣ Fundamental Theorem of Calculus - Part 2

The second part of the FTOC is more frequently used to solve definite integrals. If a function is continuous on the interval $[a,b]$ and $F$ is an antiderivative of $f$ , then

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

In other words, this theorem tells us that to find the definite integral of a function over an interval, you can evaluate its antiderivative at the upper limit of the interval and subtract the antiderivative evaluated at the lower limit. Instead of finding the integral, you can just do simple subtraction! ➖

#🪜 Steps to Solving a Definite Integral with the FTOC

Here are some steps you can follow when solving a definite integral with the fundamental theorem of calculus:

◀️ Find the antiderivative.
⬆️ Plug in the upper bound into the antiderivative.
⬇️ Plug in the lower bound into the antiderivative.
🟰 Subtract the lower bound from the upper bound and get the answer!

Now, let’s put these steps to a test and apply them to a problem!

#✏️ Solving with FTOC Part 2 Walkthrough

Evaluate the following integral.

$\int_{0}^{5}e^xdx$

According to the FTOC, to solve this definite integral, we would have to find the antiderivative of $e^x$ , then subtract the antiderivative of $0$ from the antiderivative of $5$ .

$\int_{0}^{5}e^xdx=e^5-e^0=e^5-1$

The final answer is $e^5-1$ because the antiderivative of $e^x$ is $e^x$ . Then we plugged in 5 and 0 for $x$ and got our final answer!

#🗒️ FTOC Part 2 Practice Problems

For some additional practice, complete each of these questions.

Calculate the integral of $f(x) = 3x^2$ from $a = 1$ to $b = 4$ .
Evaluate the integral $\int_{0}^{2} (5x - 2) dx$ .
Find the definite integral of $f(x)=2e^x$ from $a=0$ to $b=1$ .
Evaluate $\int_{2}^{3} \frac{1}{x} dx$ .

Before you move on to the answers, make sure you tried your best with these!

#FTOC Part 2 Question 1 Solution

First, find an antiderivative of $f(x)$ . An antiderivative of $3x^2$ is $F(x) = x^3$ . Then, use the theorem:

$\int_{1}^{4} 3x^2 dx = F(4) - F(1)$

$4^3 - 1^3 = 64 - 1 = \boxed{63}$

#FTOC Part 2 Question 2 Solution

Find the integral of $5x-2$ , which is $F(x)=\frac{5}{2}x^2 - 2x$ . Now, you can calculate:

$\int_{0}^{2} (5x - 2) dx = F(2) - F(0)$

$=\left( \frac{5}{2} \cdot 2^2 - 2 \cdot 2 \right) - \left( \frac{5}{2} \cdot 0^2 - 2 \cdot 0 \right)$

$=10 - 4 = \boxed6$

#FTOC Part 2 Question 3 Solution

The integral of $2e^x$ is $F(x)=2e^x$ itself! Therefore, we just have to apply the theorem.

$\int_{0}^{1} 2e^x dx = F(1) - F(0)$

$= 2e^1 - 2e^0 = \boxed{2e - 2}$

#FTOC Part 2 Question 4 Solution

Last but not least! The integral of $\frac{1}{x}$ is $\ln|x|$ . Therefore…

$\int_{2}^{3} \frac{1}{x} dx = \ln|3| - \ln|2|$

Great work! 👏

#📕 Closing

With practice, you will be able to evaluate any definite integrals thrown your way! The FTOC will continue to be useful throughout all integral calculus, so be sure to practice to perfection! 🌟