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 \(f \left(\right. x \left.\right)\) is continuous over an interval \(\left[\right. a , b \left]\right. ,\) and the function \(F \left(\right. x \left.\right)\) is defined by

\[F \left(\right. x \left.\right) = \int_{a}^{x} f \left(\right. t \left.\right) d t ,\]

then \(F^{'} ( x \left.\right) = f \left(\right. x \left.\right)\) over \(\left(\right. a , b \left.\right) .\)

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, \(F \left(\right. x \left.\right) ,\) as the definite integral of another function, \(f \left(\right. t \left.\right) ,\) from the point a to the point 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 x, the definite integral is a number. So the function \(F \left(\right. x \left.\right)\) returns a number (the value of the definite integral) for each value of 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.

Proof

Applying the definition of the derivative, we have

\[\begin{aligned} \\ \\ \\ F^{'} ( x \left.\right) & = \underset{h \rightarrow 0}{\text{lim}} \frac{F \left(\right. x + h \left.\right) - F \left(\right. x \left.\right)}{h} \\ \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{1}{h} \left[\right. \int_{a}^{x + h} f \left(\right. t \left.\right) d t - \int_{a}^{x} f \left(\right. t \left.\right) d t \left]\right. \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{1}{h} \left[\right. \int_{a}^{x + h} f \left(\right. t \left.\right) d t + \int_{x}^{a} f \left(\right. t \left.\right) d t \left]\right. \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{1}{h} \int_{x}^{x + h} f \left(\right. t \left.\right) d t . \end{aligned}\]

Looking carefully at this last expression, we see \(\frac{1}{h} \int_{x}^{x + h} f \left(\right. t \left.\right) d t\) is just the average value of the function \(f \left(\right. x \left.\right)\) over the interval \(\left[\right. x , x + h \left]\right. .\) Therefore, by The Mean Value Theorem for Integrals, there is some number c in \(\left[\right. x , x + h \left]\right.\) such that

\[\frac{1}{h} \int_{x}^{x + h} f \left(\right. x \left.\right) d x = f \left(\right. c \left.\right) .\]

In addition, since c is between x and x + h, c approaches x as h approaches zero. Also, since \(f \left(\right. x \left.\right)\) is continuous, we have \(\underset{h \rightarrow 0}{\text{lim}} f \left(\right. c \left.\right) = \underset{c \rightarrow x}{\text{lim}} f \left(\right. c \left.\right) = f \left(\right. x \left.\right) .\) Putting all these pieces together, we have

\[\begin{aligned} \\ F^{'} ( x \left.\right) & = \underset{h \rightarrow 0}{\text{lim}} \frac{1}{h} \int_{x}^{x + h} f \left(\right. x \left.\right) d x \\ & = \underset{h \rightarrow 0}{\text{lim}} f \left(\right. c \left.\right) \\ & = f \left(\right. x \left.\right) , \end{aligned}\]

and the proof is complete.

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Example 5.17

Finding a Derivative with the Fundamental Theorem of Calculus

Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of

\[g \left(\right. x \left.\right) = \int_{1}^{x} \frac{1}{t^{3} + 1} d t .\]

Solution

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

\[g^{'} ( x \left.\right) = \frac{1}{x^{3} + 1} .\]

Checkpoint 5.16

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

Example 5.18

Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives

Let \(F \left(\right. x \left.\right) = \int_{1}^{\sqrt{x}} \text{sin} t d t .\) Find \(F^{'} ( x \left.\right) .\)

Solution

Letting \(u \left(\right. x \left.\right) = \sqrt{x} ,\) we have \(F \left(\right. x \left.\right) = \int_{1}^{u \left(\right. x \left.\right)} \text{sin} t d t .\) Thus, by the Fundamental Theorem of Calculus and the chain rule,

\[\begin{aligned} \\ F^{'} ( x \left.\right) & = \text{sin} \left(\right. u \left(\right. x \left.\right) \left.\right) \frac{d u}{d x} \\ & = \text{sin} \left(\right. u \left(\right. x \left.\right) \left.\right) \cdot \left(\right. \frac{1}{2} x^{−1 / 2} \left.\right) \\ & = \frac{\text{sin} \sqrt{x}}{2 \sqrt{x}} . \end{aligned}\]

Checkpoint 5.17

Let \(F \left(\right. x \left.\right) = \int_{1}^{x^{3}} \text{cos} t d t .\) Find \(F^{'} ( x \left.\right) .\)

Example 5.19

Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration

Let \(F \left(\right. x \left.\right) = \int_{x}^{2 x} t^{3} d t .\) Find \(F^{'} ( x \left.\right) .\)

Solution

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

\[\begin{aligned} \\ F \left(\right. x \left.\right) & = \int_{x}^{2 x} t^{3} d t \\ & = \int_{x}^{0} t^{3} d t + \int_{0}^{2 x} t^{3} d t \\ & = − \int_{0}^{x} t^{3} d t + \int_{0}^{2 x} t^{3} d t . \end{aligned}\]

Differentiating the first term, we obtain

\[\frac{d}{d x} \left[\right. − \int_{0}^{x} t^{3} d t \left]\right. = − x^{3} .\]

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

\[\begin{aligned} \\ \frac{d}{d x} \left[\right. \int_{0}^{2 x} t^{3} d t \left]\right. & = \frac{d}{d x} \left[\right. \int_{0}^{u \left(\right. x \left.\right)} t^{3} d t \left]\right. \\ & = \left(\left(\right. u \left(\right. x \left.\right) \left.\right)\right)^{3} \frac{d u}{d x} \\ & = \left(\left(\right. 2 x \left.\right)\right)^{3} \cdot 2 \\ & = 16 x^{3} . \end{aligned}\]

Thus,

\[\begin{aligned} \\ \\ F^{'} ( x \left.\right) & = \frac{d}{d x} \left[\right. − \int_{0}^{x} t^{3} d t \left]\right. + \frac{d}{d x} \left[\right. \int_{0}^{2 x} t^{3} d t \left]\right. \\ & = − x^{3} + 16 x^{3} \\ & = 15 x^{3} . \end{aligned}\]

Checkpoint 5.18

Let \(F \left(\right. x \left.\right) = \int_{x}^{x^{2}} \text{cos} t d t .\) Find \(F^{'} ( x \left.\right) .\)

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

Theorem 5.4