5 Integration 5.2 The Definite Integral 5.4 Integration Formulas and the Net Change Theorem

5.3 The Fundamental Theorem of Calculus

Coming Concepts

•

Describe the meaning of the Mean Value Theorem for Integrals.
•

State the meaning of the Fundamental Theorem of Calculus, Part 1.
•

Use the Fundamental Theorem of Calculus, Part 1, to evaluate derivatives of integrals.
•

State the meaning of the Fundamental Theorem of Calculus, Part 2.
•

Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals.
•

Explain the relationship between differentiation and integration.

In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function. Unfortunately, so far, the only tools we have available to calculate the value of a definite integral are geometric area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section we look at some more powerful and useful techniques for evaluating definite integrals.

These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and explored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s, and it is codified in what we now call the Fundamental Theorem of Calculus, which has two parts that we examine in this section. Its very name indicates how central this theorem is to the entire development of calculus.

External Resource.

Isaac Newton’s contributions to mathematics and physics changed the way we look at the world. The relationships he discovered, codified as Newton’s laws and the law of universal gravitation, are still taught as foundational material in physics today, and his calculus has spawned entire fields of mathematics. To learn more, read a brief biography of Newton with multimedia clips.

Before we get to this crucial theorem, however, let’s examine another important theorem, the Mean Value Theorem for Integrals, which is needed to prove the Fundamental Theorem of Calculus.

5.3.1 The Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at some point in that interval. The theorem guarantees that if $\displaystyle f(x)$ is continuous, a point $\displaystyle c$ exists in an interval $\displaystyle[a,b]$ such that the value of the function at $\displaystyle c$ is equal to the average value of $\displaystyle f(x)$ over $\displaystyle[a,b]$ . We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.

Theorem 5.3 (The Mean Value Theorem for Integrals).

If $\displaystyle f(x)$ is continuous over an interval $\displaystyle[a,b]$ , then there is at least one point $\displaystyle c\in[a,b]$ such that

f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx.

(5.15)

This formula can also be stated as

\displaystyle\int_{a}^{b}f(x)\,dx=f(c)(b-a).

Proof.

Since $\displaystyle f(x)$ is continuous on $\displaystyle[a,b]$ , by the extreme value theorem (see Maxima and Minima), it assumes minimum and maximum values— $\displaystyle m$ and $\displaystyle M$ , respectively—on $\displaystyle[a,b]$ . Then, for all $\displaystyle x$ in $\displaystyle[a,b]$ , we have $\displaystyle m\leq f(x)\leq M$ . Therefore, by the Comparison Theorem), we have

\displaystyle m(b-a)\leq\int_{a}^{b}f(x)\,dx\leq M(b-a).

Dividing by $\displaystyle b-a$ gives us

\displaystyle m\leq\frac{1}{b-a}\int_{a}^{b}f(x)\,dx\leq M.

Since $\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x)\,dx$ is a number between $\displaystyle m$ and $\displaystyle M$ , and since $\displaystyle f(x)$ is continuous and assumes the values $\displaystyle m$ and $\displaystyle M$ over $\displaystyle[a,b]$ , by the Intermediate Value Theorem (see Continuity), there is a number $\displaystyle c$ over $\displaystyle[a,b]$ such that

\displaystyle f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx,

and the proof is complete.

∎

Example 5.3.1 (Finding the Average Value of a Function).

Find the average value of the function $\displaystyle f(x)=8-2x$ over the interval $\displaystyle[0,4]$ and find $\displaystyle c$ such that $\displaystyle f(c)$ equals the average value of the function over $\displaystyle[0,4]$ .

Solution: The formula states the mean value of $\displaystyle f(x)$ is given by

\displaystyle\frac{1}{4-0}\int_{0}^{4}(8-2x)\,dx.

We can see in Figure 5.31 that the function represents a straight line and forms a right triangle bounded by the $\displaystyle x$ - and $\displaystyle y$ -axes. The area of the triangle is $\displaystyle A=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)$ . We have

\displaystyle A=\frac{1}{2}(4)(8)=16.

The average value is found by multiplying the area by $\displaystyle 1/(4-0)$ . Thus, the average value of the function is

\displaystyle\frac{1}{4}\left(16\right)=4.

Set the average value equal to $\displaystyle f(c)$ and solve for $\displaystyle c$ .

	$\displaystyle 8-2c$	$\displaystyle=4$
	$\displaystyle c$	$\displaystyle=2$

At $\displaystyle c=2,f(2)=4$ .

Figure 5.31: By the Mean Value Theorem, the continuous function

\displaystyle f(x)

takes on its average value at

\displaystyle c

at least once over a closed interval.

Checkpoint 5.3.1.

Find the average value of the function $\displaystyle f(x)=\frac{x}{2}$ over the interval $\displaystyle[0,6]$ and find $\displaystyle c$ such that $\displaystyle f(c)$ equals the average value of the function over $\displaystyle[0,6]$ .

Hint: Use the procedures from Example 5.3.1 to solve the problem

Example 5.3.2 (Finding the Point Where a Function Takes on Its Average Value).

Given $\displaystyle\int_{0}^{3}x^{2}\,dx=9$ , find $\displaystyle c$ such that $\displaystyle f(c)$ equals the average value of $\displaystyle f(x)=x^{2}$ over $\displaystyle[0,3]$ .

Solution: We are looking for the value of $\displaystyle c$ such that

\displaystyle f(c)=\frac{1}{3-0}\int_{0}^{3}x^{2}\,dx=\frac{1}{3}(9)=3.

Replacing $\displaystyle f(c)$ with $\displaystyle c$ ², we have

	$\displaystyle c^{2}$	$\displaystyle=3$
	$\displaystyle c$	$\displaystyle=\pm\sqrt{3}.$

Since $\displaystyle-\sqrt{3}$ is outside the interval, take only the positive value. Thus, $\displaystyle c=\sqrt{3}$ (Figure 5.32).

A graph of the parabola $\displaystylef(x)=x^{2}$ over [-2, 3]. The area under the curve and above the x axis is shaded, and the point (sqrt(3), 3) is marked. — Figure 5.32: Over the interval $\displaystyle[0,3]$ , the function $\displaystyle f(x)=x^{2}$ takes on its average value at $\displaystyle c=\sqrt{3}$ .

Checkpoint 5.3.2.

Given $\displaystyle\int_{0}^{3}(2x^{2}-1)\,dx=15$ , find $\displaystyle c$ such that $\displaystyle f(c)$ equals the average value of $\displaystyle f(x)=2x^{2}-1$ over $\displaystyle[0,3]$ .

Hint: Use the procedures from Example 5.3.2 to solve the problem.

5.3.2 Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.

Theorem 5.4 (Fundamental Theorem of Calculus, Part 1).

If $\displaystyle f(x)$ is continuous over an interval $\displaystyle[a,b]$ , and the function $\displaystyle F(x)$ is defined by

F(x)=\int_{a}^{x}f(t)\,dt,

(5.16)

then $\displaystyle F^{\prime}(x)=f(x)$ over $\displaystyle[a,b]$ .

Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, $\displaystyle F(x)$ , as the definite integral of another function, $\displaystyle f(t)$ , from the point $\displaystyle a$ to the point $\displaystyle x$ . At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of $\displaystyle x$ , the definite integral is a number. So the function $\displaystyle F(x)$ returns a number (the value of the definite integral) for each value of $\displaystyle x$ .

Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.

Proof.

Applying the definition of the derivative, we have

	$\displaystyle F^{\prime}(x)$	$\displaystyle=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}$
		$\displaystyle=\lim_{h\to 0}\frac{1}{h}\left[\int_{a}^{x+h}f(t)\,dt-\int_{a}^{x% }f(t)\,dt\right]$
		$\displaystyle=\lim_{h\to 0}\frac{1}{h}\left[\int_{a}^{x+h}f(t)\,dt+\int_{x}^{a% }f(t)\,dt\right]$
		$\displaystyle=\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}f(t)\,dt.$

Looking carefully at this last expression, we see $\displaystyle\frac{1}{h}\int_{x}^{x+h}f(t)\,dt$ is just the average value of the function $\displaystyle f(x)$ over the interval $\displaystyle[x,x+h]$ . Therefore, by Theorem 5.3, there is some number $\displaystyle c$ in $\displaystyle[x,x+h]$ such that

\displaystyle\frac{1}{h}\int_{x}^{x+h}f(x)\,dx=f(c).

In addition, since $\displaystyle c$ is between $\displaystyle x$ and $\displaystyle x$ + $\displaystyle h$ , $\displaystyle c$ approaches $\displaystyle x$ as $\displaystyle h$ approaches zero. Also, since $\displaystyle f(x)$ is continuous, we have $\displaystyle\lim_{h\to 0}f(c)=\lim_{c\to x}f(c)=f(x)$ . Putting all these pieces together, we have

	$\displaystyle F^{\prime}(x)$	$\displaystyle=\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}f(x)\,dx$
		$\displaystyle=\lim_{h\to 0}f(c)$
		$\displaystyle=f(x),$

and the proof is complete.

∎

Example 5.3.3 (Finding a Derivative with the Fundamental Theorem of Calculus).

Use the Theorem 5.4 to find the derivative of

\displaystyle g(x)=\int_{1}^{x}\frac{1}{t^{3}+1}\,dt.

Solution: According to the Fundamental Theorem of Calculus, the derivative is given by

\displaystyle g^{\prime}(x)=\frac{1}{x^{3}+1}.

Checkpoint 5.3.3.

Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of $\displaystyle g(r)=\int_{0}^{r}\sqrt{x^{2}+4}\,dx$ .

Hint: Follow the procedures from Example 5.3.3 to solve the problem.

Example 5.3.4 (Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives).

Let $\displaystyle F(x)=\int_{1}^{\sqrt{x}}\sin t\,dt$ . Find $\displaystyle F^{\prime}(x)$ .

Solution: Letting $\displaystyle u(x)=\sqrt{x}$ , we have $\displaystyle F(x)=\int_{1}^{u(x)}\sin t\,dt$ . Thus, by the Fundamental Theorem of Calculus and the chain rule,

	$\displaystyle F^{\prime}(x)$	$\displaystyle=\sin(u(x))\frac{du}{dx}$
		$\displaystyle=\sin(u(x))\cdot\left(\frac{1}{2}x^{-1/2}\right)$
		$\displaystyle=\frac{\sin\sqrt{x}}{2\sqrt{x}}.$

Checkpoint 5.3.4.

Let $\displaystyle F(x)=\int_{1}^{x^{3}}\cos t\,dt$ . Find $\displaystyle F^{\prime}(x)$ .

Hint: Use the chain rule to solve the problem.

Example 5.3.5 (Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration).

Let $\displaystyle F(x)=\int_{x}^{2x}t^{3}\,dt$ . Find $\displaystyle F^{\prime}(x)$ .

Solution: We have $\displaystyle F(x)=\int_{x}^{2x}t^{3}\,dt$ . Both limits of integration are variable, so we need to split this into two integrals. We get

	$\displaystyle F(x)$	$\displaystyle=\int_{x}^{2x}t^{3}\,dt$
		$\displaystyle=\int_{x}^{0}t^{3}\,dt+\int_{0}^{2x}t^{3}\,dt$
		$\displaystyle=-\int_{0}^{x}t^{3}\,dt+\int_{0}^{2x}t^{3}\,dt.$

Differentiating the first term, we obtain

\displaystyle\frac{d}{dx}\left[-\int_{0}^{x}t^{3}\,dt\right]=-x^{3}.

Differentiating the second term, we first let $\displaystyle u(x)=2x$ . Then,

	$\displaystyle\frac{d}{dx}\left[\int_{0}^{2x}t^{3}\,dt\right]$	$\displaystyle=\frac{d}{dx}\left[\int_{0}^{u(x)}t^{3}\,dt\right]$
		$\displaystyle=(u(x))^{3}\frac{du}{dx}$
		$\displaystyle=(2x)^{3}\cdot 2$
		$\displaystyle=16x^{3}.$

Thus,

	$\displaystyle F^{\prime}(x)$	$\displaystyle=\frac{d}{dx}\left[-\int_{0}^{x}t^{3}\,dt\right]+\frac{d}{dx}% \left[\int_{0}^{2x}t^{3}\,dt\right]$
		$\displaystyle=-x^{3}+16x^{3}$
		$\displaystyle=15x^{3}.$

Checkpoint 5.3.5.

Let $\displaystyle F(x)=\int_{x}^{x^{2}}\cos t\,dt$ . Find $\displaystyle F^{\prime}(x)$ .

Hint: Use the procedures from Example 5.3.5 to solve the problem.

5.3.3 Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.

After finding approximate areas by adding the areas of $\displaystyle n$ rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.

Theorem 5.5 (The Fundamental Theorem of Calculus, Part 2).

If $\displaystyle f$ is continuous over the interval $\displaystyle[a,b]$ and $\displaystyle F(x)$ is any antiderivative of $\displaystyle f(x)$ , then

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

(5.17)

We often see the notation $\displaystyle{F(x)}\,\bigg{|}_{a}^{b}$ to denote the expression $\displaystyle F(b)-F(a)$ . We use this vertical bar and associated limits $\displaystyle a$ and $\displaystyle b$ to indicate that we should evaluate the function $\displaystyle F(x)$ at the upper limit (in this case, $\displaystyle b$ ), and subtract the value of the function $\displaystyle F(x)$ evaluated at the lower limit (in this case, $\displaystyle a$ ).

The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.

Proof.

Let $\displaystyle P=\{x_{i}\},i=0,1,\ldots,n$ be a regular partition of $\displaystyle[a,b]$ . Then, we can write

	$\displaystyle F(b)-F(a)$	$\displaystyle=F(x_{n})-F(x_{0})$
		$\displaystyle=[F(x_{n})-F(x_{n-1})]+[F(x_{n-1})-F(x_{n-2})]+\ldots+[F(x_{1})-F% (x_{0})]$
		$\displaystyle=\sum_{i=1}^{n}\left[F(x_{i})-F(x_{i-1})\right].$

Now, we know $\displaystyle F$ is an antiderivative of $\displaystyle f$ over $\displaystyle[a,b]$ , so by the Mean Value Theorem for $\displaystyle i=0,1,\ldots,n$ we can find $\displaystyle c_{i}$ in $\displaystyle[x_{i-1},x_{i}]$ such that

\displaystyle F(x_{i})-F(x_{i-1})=F^{\prime}(c_{i})(x_{i}-x_{i-1})=f(c_{i})% \Delta x.

Then, substituting into the previous equation, we have

\displaystyle F(b)-F(a)=\sum_{i=1}^{n}f(c_{i})\Delta x.

Taking the limit of both sides as $\displaystyle n\to\infty$ , we obtain

	$\displaystyle F(b)-F(a)$	$\displaystyle=\lim_{n\to\infty}\sum_{i=1}^{n}f(c_{i})\Delta x$
		$\displaystyle=\int_{a}^{b}f(x)\,dx.$

∎

Example 5.3.6 (Evaluating an Integral with the Fundamental Theorem of Calculus).

Use Theorem 5.5 to evaluate

\displaystyle\int_{-2}^{2}(t^{2}-4)\,dt.

Solution: Recall the power rule for Antiderivatives:

\displaystyle\text{ if }y=x^{n},\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C.

Use this rule to find the antiderivative of the function and then apply the theorem. We have

	$\displaystyle\int_{-2}^{2}(t^{2}-4)\,dt$	$\displaystyle=\frac{t^{3}}{3}-{4t\|}_{-2}^{2}$
		$\displaystyle=\left[\frac{(2)^{3}}{3}-4(2)\right]-\left[\frac{(-2)^{3}}{3}-4% \left(-2\right)\right]$
		$\displaystyle=\left(\frac{8}{3}-8\right)-\left(-\frac{8}{3}+8\right)$
		$\displaystyle=\frac{8}{3}-8+\frac{8}{3}-8$
		$\displaystyle=\frac{16}{3}-16$
		$\displaystyle=-\frac{32}{3}.$

Analysis: Notice that we did not include the “+ $\displaystyle C$ ” term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose the antiderivative with $\displaystyle C=0$ . If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.

The region of the area we just calculated is depicted in Figure 5.33. Note that the region between the curve and the $\displaystyle x$ -axis is all below the $\displaystyle x$ -axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.

The graph of the parabola $\displaystylef(t)=t^{2}–4$ over [-4, 4]. The area above the curve and below the x axis over [-2, 2] is shaded. — Figure 5.33: The evaluation of a definite integral can produce a negative value, even though area is always positive.

Example 5.3.7 (Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2).

Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:

\displaystyle\int_{1}^{9}\frac{x-1}{\sqrt{x}}\,dx.

Solution: First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:

\displaystyle\int_{1}^{9}\frac{x-1}{x^{1/2}}\,dx=\int_{1}^{9}\left(\frac{x}{x^% {1/2}}-\frac{1}{x^{1/2}}\right)\,dx.

Use the properties of exponents to simplify:

\displaystyle\int_{1}^{9}\left(\frac{x}{x^{1/2}}-\frac{1}{x^{1/2}}\right)\,dx=% \int_{1}^{9}(x^{1/2}-x^{-1/2})\,dx.

Now, integrate using the power rule:

	$\displaystyle\int_{1}^{9}(x^{1/2}-x^{-1/2})\,dx$	$\displaystyle={\left(\frac{x^{3/2}}{\frac{3}{2}}-\frac{x^{1/2}}{\frac{1}{2}}% \right)}\,\bigg{\|}_{1}^{9}$
		$\displaystyle=\left[\frac{(9)^{3/2}}{\frac{3}{2}}-\frac{(9)^{1/2}}{\frac{1}{2}% }\right]-\left[\frac{(1)^{3/2}}{\frac{3}{2}}-\frac{(1)^{1/2}}{\frac{1}{2}}\right]$
		$\displaystyle=\left[\frac{2}{3}\left(27\right)-2(3)\right]-\left[\frac{2}{3}(1% )-2(1)\right]$
		$\displaystyle=18-6-\frac{2}{3}+2$
		$\displaystyle=\frac{40}{3}.$

See Figure 5.34.

The graph of the function f(x) = (x-1) / sqrt(x) over
[0,9]. The area under the graph over [1,9] is shaded. — Figure 5.34: The area under the curve from $\displaystyle x=1$ to $\displaystyle x=9$ can be calculated by evaluating a definite integral.

Checkpoint 5.3.6.

Use Theorem 5.5 to evaluate $\displaystyle\int_{1}^{2}x^{-4}\,dx$ .

Hint: Use the power rule.

Example 5.3.8 (A Roller-Skating Race).

James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after 5 sec wins a prize. If James can skate at a velocity of $\displaystyle f(t)=5+2t$ ft/sec and Kathy can skate at a velocity of $\displaystyle g(t)=10+\cos\left(\frac{\pi}{2}t\right)$ ft/sec, who is going to win the race?

Solution: We need to integrate both functions over the interval $\displaystyle[0,5]$ and see which value is bigger. For James, we want to calculate

\displaystyle\int_{0}^{5}(5+2t)\,dt.

Using the power rule, we have

	$\displaystyle\int_{0}^{5}(5+2t)\,dt$	$\displaystyle={(5t+t^{2})}\,\bigg{\|}_{0}^{5}$
		$\displaystyle=(25+25)=50.$

Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate

\displaystyle\int_{0}^{5}10+\cos\left(\frac{\pi}{2}t\right)\,dt.

We know $\displaystyle\sin t$ is an antiderivative of $\displaystyle\cos t$ , so it is reasonable to expect that an antiderivative of $\displaystyle\cos\left(\frac{\pi}{2}t\right)$ would involve $\displaystyle\sin\left(\frac{\pi}{2}t\right)$ . However, when we differentiate $\displaystyle\sin\left(\frac{\pi}{2}t\right)$ , we get $\displaystyle\frac{\pi}{2}\cos\left(\frac{\pi}{2}t\right)$ as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain

	$\displaystyle\int_{0}^{5}10+\cos\left(\frac{\pi}{2}t\right)\,dt$	$\displaystyle={\left(10t+\frac{2}{\pi}\sin\left(\frac{\pi}{2}t\right)\right)}% \,\bigg{\|}_{0}^{5}$
		$\displaystyle=\left(50+\frac{2}{\pi}\right)-\left(0-\frac{2}{\pi}\sin 0\right)$
		$\displaystyle\approx 50.6.$

Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!

Checkpoint 5.3.7.

Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec. Does this change the outcome?

Hint: Change the limits of integration from those in Example 5.3.8.

Key Concepts

•

The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value $\displaystyle c$ such that $\displaystyle f(c)$ equals the average value of the function. See Theorem 5.3.
•

The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. See Theorem 5.4.
•

The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. The total area under a curve can be found using this formula. See Theorem 5.5.

Key Equations

•

Mean Value Theorem for Integrals

If $\displaystyle f(x)$ is continuous over an interval $\displaystyle[a,b]$ , then there is at least one point $\displaystyle c\in[a,b]$ such that $\displaystyle f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx$ .
•

Fundamental Theorem of Calculus Part 1

If $\displaystyle f(x)$ is continuous over an interval $\displaystyle[a,b]$ , and the function $\displaystyle F(x)$ is defined by $\displaystyle F(x)=\int_{a}^{x}f(t)\,dt$ , then $\displaystyle F^{\prime}(x)=f(x)$ .
•

Fundamental Theorem of Calculus Part 2

If $\displaystyle f$ is continuous over the interval $\displaystyle[a,b]$ and $\displaystyle F(x)$ is any antiderivative of $\displaystyle f(x)$ , then $\displaystyle\int_{a}^{b}f(x)\,dx=F(b)-F(a)$ .

Glossary

fundamental theorem of calculus: the theorem, central to the entire development of calculus, that establishes the relationship between differentiation and integration
fundamental theorem of calculus, part 1: uses a definite integral to define an antiderivative of a function
fundamental theorem of calculus, part 2: (also, evaluation theorem) we can evaluate a definite integral by evaluating the antiderivative of the integrand at the endpoints of the interval and subtracting
mean value theorem for integrals: guarantees that a point $\displaystyle c$ exists such that $\displaystyle f(c)$ is equal to the average value of the function