# The Fundamental Theorem of Calculus

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Calculus $\longrightarrow$
Understanding Integration $\longrightarrow$

Objectives
• Relate functions and the area under their curve using Antiderivatives and the Fundamental Theorem of Calculus
• Use the FTC to evaluate definite integrals
• Understand Part 2 of the FTC, involving simultaneous derivatives and integrals
Lesson Description

The definite integral of a known function does not require a graph or computer, but rather can be expressed algebraically using the antiderivatives and the Fundamental Theorem of Calculus. This lesson shows us how to execute on definite integrals of known functions, which we will continue practicing as we learn more about specific function families' antiderivative processes.

Practice Problems

Practice problems and worksheet coming soon!

## The Integral Payoff

Combining what we've learned about what antiderivatives » are, as well as the properties and meaning of definite integrals », we can state the Fundamental Theorem of Calculus (FTC for short), which will guide our function-specific work on definite integrals.
The Fundamental Theorem of CalculusLet $F(x)$ be the antiderivative of $f(x)$, such that$$\int f(x) \, dx = F(x) + C$$It follows that$$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$$
In words, this means that the definite integral of function $f(x)$ is equal to the function's antiderivative evaluated at the upper limit, minus the function's antiderivative evaluated at the lower limit.Evaluation NotationDuring the in-between step of evaluating a definite integral after we have figured out the proper antiderivative to use but before we plug in the arithmetic, it is common for us to use a vertical bar on the right-hand side of the antiderivative and write the limits of integration. This notation is shorthand for "we are about to evaluate this expression at these limits".For example, if the antiderivative of $\cos(x)$ is $\sin(x) + C$ then:$$\int_{0}^{\pi/2} \cos(x) \, dx = \sin(x) \Bigg\rvert_{0}^{\pi/2}$$We can't use the $\int$ symbol for this purpose because the $\int$ symbol implies you haven't yet done the antiderivative step yet. Also, we don't want to jump right in an plug in because it's important to write the antiderivative expression on its own first, to not only help ourselves keep track, but also to show the grader that we used the correct antiderivative. Now we can finish the problem.$$\sin(x) \Bigg\rvert_{0}^{\pi/2} = \sin(\pi/2) - \sin(0)$$$$=1 \;\; \blacksquare$$

## FTC in Action

As we said, the FTC is used to evaluate definite integrals. Routinely, you'll have to determine the function's antiderivative before plugging in the integration limits. Since we will begin to practice antiderivatives of specific function families in the next few lessons, for now we'll practice with given information.

Example 1Given that $\displaystyle \int 2x \, dx = x^2 + C$, determine the value of$$\int_{1}^{4} 2x \, dx$$$\blacktriangleright$ According to the FTC:$$\int_{1}^{4} 2x \, dx = x^2 \Bigg\rvert_{1}^{4}$$$$=(4)^2 - (1)^2 = 15$$

Example 2Given that$$\int \frac{x}{\sqrt{x^2-3}} \; dx = \sqrt{x^2 -3} + C$$Calculate$$\int_{2}^{\sqrt{12}} \frac{x}{\sqrt{x^2-3}} \; dx$$$\blacktriangleright$ The definite integral evaluates to the antiderivative, evaluated at each limit, subtracted from one another (upper limit minus lower limit).$$\int_{2}^{\sqrt{12}} \frac{x}{\sqrt{x^2-3}} \; dx = \sqrt{x^2 -3} \Bigg\rvert_{2}^{/sqrt{12}}$$$$= \left( \sqrt{\left(\sqrt{12}\right)^2 - 3} \right) - \left( \sqrt{ 2^2 - 3} \right)$$$$=\sqrt{9} - \sqrt{1} = 2$$$$\blacksquare$$

## FTC Version 2

The reason that mathematicians named this principle the Fundamental Theory of Calculus is that it captures the exact nature of the relationship between derivatives and integrals. There is another way to state this relationship by reordering the operations, and most teachers and texts refer to it as the Second Fundamental Theorem of Calculus, or the Fundamental Theorem of Calculus version 2, or alternate, or alias, etc.I typically refer to it as FTC 2.
Theorem: FTC 2Let $x$ be the variable of a function $f(x)$, $c$ be a real number constant, and $t$ be a variable.Then, the derivative of the integral of function $f$ is equal to function $f$. In this way, the derivative and integral operations are inverses. Symbolically:$$\frac{d}{dx} \; \int_{c}^{x} f(t) \, dt = f(x)$$
Let's see why FTC 2 makes sense with a proving example (not a rigorous proof however).

Example 3Prove the FTC version 2 to be true by showing that$$\frac{d}{dx} \, \int_{3}^{x} 3t^2 \, dt = 3x^2$$given that you are told$$\int 3x^2 \, dx = x^3 + C$$$\blacktriangleright$ Let's start by evaluating the definite integral on its own.$$\int_{3}^{x} 3t^2 \, dt = t^3 \Bigg\rvert_{3}^{x}$$$$= x^3 - 3^3 = x^3 - 27$$Now, plug this result into the original problem:$$\frac{d}{dx} \, \int_{3}^{x} 3t^2 \, dt$$$$\longrightarrow \frac{d}{dx} \;\; \left[ x^3 - 27 \right]$$$$= 3x^2$$$$\blacksquare$$

This version of the derivative - integral relationship is slightly less useful, but lends itself well to a particular type of popular quiz question. The idea is, since you get back what you started with, you don't even need to know the antiderivative in these cases.

Example 4Using the FTC, determine the value of$$\frac{d}{dx} \, \int_{1}^{x} e^{t^2} \, dt$$$\blacktriangleright$ The idea here is that we cannot easily find the antiderivative of $e^{t^2}$ (and not only because we haven't yet learned much about integrals of exponentials - this one is actually difficult). However, as FTC version 2 states, whatever $x$ expression the antiderivative is, we are going to take its derivative anyway which will leave us with the original function. Therefore,$$\frac{d}{dx} \, \int_{1}^{x} e^{t^2} \, dt = e^{x^2}$$

## Mr. Math Makes It Mean

Backwards FTC 2A small twist you'll see on quizzes is having the $x$ integration limit as the lower bound. To adjust for this, switch the limits » of integration. It introduces a factor of $-1$ but allows you to proceed with FTC 2.

Example 5Evaluate$$\frac{d}{dx} \int_{x}^{2} \cos\left( t^2 \right) \, dt$$$\blacktriangleright$ By the property of switching limits of integration:$$\int_{x}^{2} \cos\left( t^2 \right) \, dt = -\int_{2}^{x} \cos\left( t^2 \right) \, dt$$Now, we can directly apply FTC 2.$$frac{d}{dx} - \int_{2}^{x} \cos\left( t^2 \right) \, dt$$$$=-\cos\left( x^2 \right)$$

## Put It To The Test

Example 7Given that$$\int x^5 \, dx = \frac{x^6}{6} + C$$evaluate$$\int_{1}^{6} x^5 \, dx$$
Show solution
$\blacktriangleright$ Using FTC, we need to plug in the upper and lower integration limits into the antiderivative, and subtract the results.$$\int_{1}^{6} x^5 \, dx = \frac{x^6}{6} \; \Bigg\rvert_{1}^{6}$$$$= \frac{(6)^6}{6} - \frac{(1)^6}{6} = \frac{46655}{6}$$Typically you would have access to a graphing calculator for arithmetic that involves fractions.$$\blacksquare$$

Example 8Given that$$\int 3x^4 \, dx = \frac{3x^5}{5} + C$$evaluate$$\int_{-5}^{-3} 2x^4 \, dx$$
Show solution
$\blacktriangleright$ Using the rules that govern constant coefficients of integrals, we can take the given antiderivative and multiply by $2/3$:$$\frac{2}{3} \cdot \int 3x^4 \, dx = \frac{2}{3} \cdot \frac{3x^5}{5} + C$$$$\longrightarrow \int 2x^4 \, dx = \frac{2x^5}{5} + C$$Now, apply the FTC.$$\int_{-5}^{-3} 2x^4 \, dx = \frac{2x^5}{5} \Bigg\rvert_{-5}^{-3}$$$$=\left( \frac{2(-3)^5}{5} \right) - \left( \frac{2(-5)^5}{5} \right)$$$$=\frac{5764}{5}$$Note that for simplicity, many students find it helpful to factor constant fractions out from the evaluation step. For example, we could have expressed this instead as$$\frac{2x^5}{5} \; \Bigg\rvert_{-5}^{-3} = \frac{2}{5} \cdot \bigg[ x^5 \Bigg\rvert_{-5}^{-3}$$$$=\frac{2}{5} \cdot \left( (-3)^5 - (-5)^5 \right)$$$$=\frac{5764}{5}$$$$\blacksquare$$

Example 9Given that$$\int \sin(2x) \, dx = \frac{-\cos(2x)}{2} + C$$determine$$\int_{\pi/6}^{2\pi/3} \sin{2x} \, dx$$
Show solution
$\blacktriangleright$ Let's evaluate the antiderivative at each limit and subtract. The fun part here is remembering how to evaluate trig functions, which we have to remember from Pre-Calculus.According to the FTC,$$\int_{\pi/6}^{2\pi/3} \sin{2x} \, dx = \frac{-\cos(2x)}{2} \; \Bigg\rvert_{\pi/6}^{2\pi/3}$$$$= -\frac{1}{2} \, \left( \cos\left( 2 \cdot \frac{2\pi}{3} \right) - \cos\left( 2 \cdot \frac{\pi}{6} \right) \right)$$$$= -\frac{1}{2} \, \left( \cos\left( \frac{4\pi}{3} \right) - \cos\left( \frac{\pi}{3} \right) \right)$$$$= -\frac{1}{2} \, \left( \left( - \frac{1}{2} \right) - \left(\frac{1}{2} \right) \right)$$$$= \frac{1}{2}$$$$\blacksquare$$

Example 10Evaluate.$$\frac{d}{dx} \; \int_{x}^{7} \sin\left(\sqrt{t}\right) \, dt$$
Show solution
$\blacktriangleright$ First, we must switch the integration limits to match the FTC form:$$\int_{x}^{7} \sin\left(\sqrt{t}\right) \, dt = - \int_{7}^{x} \sin\left(\sqrt{t}\right) \, dt$$Now FTC 2 can apply.$$\frac{d}{dx} - \int_{7}^{x} \sin\left(\sqrt{t}\right) \, dt$$$$= \sin\left(\sqrt{x}\right)$$$$\blacksquare$$

Lesson Takeaways
• Understand the mechanics involved in evaluating definite integrals according to the FTC
• Know conceptually how definite integrals and area bound by the function are related
• Use the secondary version of FTC to answer questions that involve derivatives of integrals or vice versa
• Be able to solve graph-based problems by applying FTC
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