Grade 12

Advanced

8 min

Lesson 6 of 12

The Definite Integral and FTC

Not Started

Connect the concepts of differentiation and integration through the Fundamental Theorem of Calculus.

Imagine you are tracking a rocket's velocity. If you know its speed at every microsecond, could you calculate its exact position 100 miles away without ever looking at a GPS? The Fundamental Theorem of Calculus is the 'secret bridge' that makes this possible.

Learning Objectives

1.Evaluate definite integrals using the Fundamental Theorem of Calculus (Part 2)
2.Interpret the definite integral as the net signed area under a curve
3.Apply the Mean Value Theorem for Integrals to calculate the average value of a function

The Fundamental Theorem of Calculus (FTC)

The Fundamental Theorem of Calculus is the crown jewel of mathematics. It connects two seemingly unrelated concepts: the derivative (slope) and the integral (area). Specifically, Part 2 of the theorem states that if

f

is continuous on

[a, b]

and

F

is any antiderivative of

f

, then:

\int_a^b f(x) dx = F(b) - F(a)

This means we no longer need to sum infinite rectangles to find area; we simply find the antiderivative at the boundaries and subtract. This process represents the net change of the quantity over the interval.

Evaluate the definite integral:

\int_1^3 x^2 dx

1. Find the antiderivative of $f(x) = x^2$ . Using the power rule, $F(x) = \frac{x^3}{3}$ . 2. Apply the FTC formula: $F(3) - F(1)$ . 3. Substitute the values: $(\frac{3^3}{3}) - (\frac{1^3}{3}) = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$ .

Quick Check

If the antiderivative of a function is $F(x) = \sin(x)$ , what is the value of the integral from $0$ to $\pi$ ?

Use the formula

F(b) - F(a)

Answer

0. Because $F(\pi) - F(0) = \sin(\pi) - \sin(0) = 0 - 0 = 0$ .

Net Signed Area vs. Total Area

The definite integral $\int_a^b f(x) dx$ calculates the net signed area. This means areas above the $x$ -axis are counted as positive, while areas below the $x$ -axis are counted as negative. If a function represents velocity, the definite integral gives the displacement (net change in position). If you want the total area (or total distance traveled), you must integrate the absolute value of the function, $|f(x)|$ , which effectively 'flips' the negative sections to be positive.

Find the net signed area of $f(x) = x$ from $x = -2$ to $x = 2$ .

1. Set up the integral: $\int_{-2}^2 x dx$ . 2. Find the antiderivative: $F(x) = \frac{x^2}{2}$ . 3. Evaluate: $F(2) - F(-2) = \frac{2^2}{2} - \frac{(-2)^2}{2} = 2 - 2 = 0$ . 4. Interpretation: The area from $-2$ to $0$ is $-2$ , and the area from $0$ to $2$ is $+2$ . They cancel out!

Quick Check

True or False: A definite integral can result in a negative number.

Answer

True. If more of the function's area lies below the x-axis than above it, the result is negative.

The Mean Value Theorem for Integrals

How do we find the 'average' height of a curve? Since a curve has infinitely many points, we can't just add them up. The Mean Value Theorem (MVT) for Integrals provides the solution. It states that for a continuous function on

[a, b]

, there exists at least one point

c

in the interval such that:

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

The value

f(c)

is known as the average value of the function. Geometrically, it's the height of a rectangle with width

(b-a)

that has the exact same area as the region under the curve.

Find the average value of $f(x) = 3x^2$ on the interval $[0, 2]$ .

1. Identify $a=0$ and $b=2$ . The width is $b - a = 2$ . 2. Calculate the integral: $\int_0^2 3x^2 dx = [x^3]_0^2 = 8 - 0 = 8$ . 3. Divide by the width: $f_{avg} = \frac{1}{2} \times 8 = 4$ . 4. Challenge: Find $c$ such that $f(c) = 4$ . Solve $3c^2 = 4 \rightarrow c^2 = \frac{4}{3} \rightarrow c = \sqrt{\frac{4}{3}} \approx 1.15$ .

Key Takeaways

1The Fundamental Theorem of Calculus (Part 2) allows us to evaluate integrals using antiderivatives: $F(b) - F(a)$ .
2Definite integrals represent 'net signed area'; regions below the x-axis subtract from the total.
3The average value of a function over $[a, b]$ is calculated as $\frac{1}{b-a} \int_a^b f(x) dx$ .

Quick Check

Multiple Choice

What is the value of $\int_1^2 6x^2 dx$ ?

Multiple Choice

If $\int_a^b f(x) dx = 0$ , what must be true?

True/False

The Mean Value Theorem for Integrals guarantees that a function will reach its average value at least once in a closed interval.

What's Next?

Review Tomorrow

In 24 hours, try to write down the FTC Part 2 formula and explain the difference between 'displacement' and 'total distance' using integrals.

Practice Activity

Find a graph of a sine wave from $0$ to $2\pi$ . Predict its definite integral value based on symmetry, then calculate it using the FTC to verify.

Browse

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The Definite Integral and FTC

Learning Objectives

The Fundamental Theorem of Calculus (FTC)

Net Signed Area vs. Total Area

The Mean Value Theorem for Integrals

Key Takeaways

Quick Check

What's Next?

The Definite Integral and FTC

Learning Objectives

The Fundamental Theorem of Calculus (FTC)

Net Signed Area vs. Total Area

The Mean Value Theorem for Integrals

Key Takeaways

Quick Check

What's Next?