The Fundamental Theorem of Calculus (HSC SSCE Mathematics Advanced): Revision Notes

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The Fundamental Theorem of Calculus

Introduction

The fundamental theorem of calculus provides a powerful method for evaluating definite integrals. This theorem connects two major concepts in calculus: differentiation (finding rates of change) and integration (finding areas). The connection between these seemingly different processes is remarkable and forms the foundation for much of calculus.

What is a primitive?

A primitive (also called an anti-derivative) is a function that reverses the process of differentiation.

infoNote

Definition: A function F(x)F(x) is called a primitive of a function f(x)f(x) if differentiating F(x)F(x) gives us f(x)f(x).

In mathematical notation: F(x)F(x) is a primitive of f(x)f(x) if F′(x)=f(x)F'(x) = f(x)

Finding primitives of power functions

For any power function xnx^n where n≠−1n \neq -1, we can find the primitive using this rule:

If dydx=xn\frac{dy}{dx} = x^n, then y=xn+1n+1+Cy = \frac{x^{n+1}}{n+1} + C

where CC is a constant.

Memory aid: "Increase the index by 1 and divide by the new index."

lightbulbExample

Finding a Primitive of a Power Function

If we want the primitive of x2x^2:

  • Increase the power from 2 to 3
  • Divide by the new power (3)
  • The primitive is x33+C\frac{x^3}{3} + C
chatImportant

This rule does not work when n=−1n = -1 because it would require dividing by zero, which is undefined. We will learn how to handle this special case when studying logarithmic functions.

The fundamental theorem of calculus

chatImportant

Statement of the Theorem:

Let f(x)f(x) be a function that is continuous in a closed interval \[a,b]\[a, b]. Then:

∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a)

where F(x)F(x) is any primitive of f(x)f(x).

What this means: To evaluate a definite integral, find any primitive of the function, substitute the upper limit, substitute the lower limit, and then subtract the second result from the first.

This theorem tells us that finding areas under curves and finding tangents to curves are inverse operations, even though this connection is not immediately obvious.

How to use the fundamental theorem

Notation

The standard way to show this working is to use square brackets:

∫abf(x) dx=[F(x)]ab=F(b)−F(a)\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)

The lower limit aa is written as a subscript and the upper limit bb as a superscript after the square bracket containing the primitive.

Basic examples

lightbulbExample

Evaluate ∫022x dx\int_0^2 2x \, dx

Step 1: Find a primitive of 2x2x

A primitive of 2x2x is x2x^2 (because differentiating x2x^2 gives 2x2x)

Step 2: Write in square brackets with limits

∫022x dx=[x2]02\int_0^2 2x \, dx = [x^2]_0^2

Step 3: Substitute upper limit (2), then lower limit (0), and subtract

=22−02= 2^2 - 0^2

=4−0= 4 - 0

=4= 4

Verification: This represents the area of a triangle with base 2 and height 4.

Area =12×2×4=4= \frac{1}{2} \times 2 \times 4 = 4 ✓

lightbulbExample

Evaluate ∫24(2x−3) dx\int_2^4 (2x - 3) \, dx

Solution:

A primitive of 2x−32x - 3 is x2−3xx^2 - 3x (take the primitive of each term)

∫24(2x−3) dx=[x2−3x]24\int_2^4 (2x - 3) \, dx = [x^2 - 3x]_2^4

=(16−12)−(4−6)= (16 - 12) - (4 - 6)

=4−(−2)= 4 - (-2)

=6= 6

This represents the area of a trapezium with parallel sides of length 1 and 5, and width 2.

Area =1+52×2=6= \frac{1 + 5}{2} \times 2 = 6 ✓

Working with polynomial functions

Evaluate these integrals:

a) ∫01x2 dx\int_0^1 x^2 \, dx

Solution:

Increase the index from 2 to 3, then divide by 3

∫01x2 dx=[x33]01\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1

=13−0= \frac{1}{3} - 0

=13= \frac{1}{3}

b) ∫−22(x3+8) dx\int_{-2}^{2} (x^3 + 8) \, dx

Solution:

Take the primitive of each term

∫−22(x3+8) dx=[x44+8x]−22\int_{-2}^{2} (x^3 + 8) \, dx = \left[\frac{x^4}{4} + 8x\right]_{-2}^{2}

Substitute x=2x = 2: 164+16=4+16=20\frac{16}{4} + 16 = 4 + 16 = 20

Substitute x=−2x = -2: 164−16=4−16=−12\frac{16}{4} - 16 = 4 - 16 = -12

=20−(−12)= 20 - (-12)

=32= 32

c) ∫04(25−x2) dx\int_0^4 (25 - x^2) \, dx

Solution:

∫04(25−x2) dx=[25x−13x3]04\int_0^4 (25 - x^2) \, dx = \left[25x - \frac{1}{3}x^3\right]_0^4

=(100−643)−(0−0)= \left(100 - \frac{64}{3}\right) - (0 - 0)

Note: Never omit substituting 0, even though it often equals zero

=100−643= 100 - \frac{64}{3}

=300−643= \frac{300 - 64}{3}

=2363= \frac{236}{3}

=7823= 78\frac{2}{3}

chatImportant

When the primitive contains two or more terms, always use brackets when substituting the upper and lower limits. This prevents errors in calculations.

Expanding brackets in the integrand

chatImportant

Since there is no product rule for integration, we must expand brackets before finding the primitive.

lightbulbExample

Expand the brackets then evaluate:

a) ∫16x(x+1) dx\int_1^6 x(x + 1) \, dx

Solution:

First expand: x(x+1)=x2+xx(x + 1) = x^2 + x

∫16x(x+1) dx=∫16(x2+x) dx\int_1^6 x(x + 1) \, dx = \int_1^6 (x^2 + x) \, dx

=[x33+x22]16= \left[\frac{x^3}{3} + \frac{x^2}{2}\right]_1^6

=(2163+362)−(13+12)= \left(\frac{216}{3} + \frac{36}{2}\right) - \left(\frac{1}{3} + \frac{1}{2}\right)

=(72+18)−(26+36)= (72 + 18) - \left(\frac{2}{6} + \frac{3}{6}\right)

=90−56= 90 - \frac{5}{6}

=8916= 89\frac{1}{6}

b) ∫03(x−4)(x−6) dx\int_0^3 (x - 4)(x - 6) \, dx

Solution:

Expand: (x−4)(x−6)=x2−10x+24(x - 4)(x - 6) = x^2 - 10x + 24

∫03(x−4)(x−6) dx=∫03(x2−10x+24) dx\int_0^3 (x - 4)(x - 6) \, dx = \int_0^3 (x^2 - 10x + 24) \, dx

=[x33−5x2+24x]03= \left[\frac{x^3}{3} - 5x^2 + 24x\right]_0^3

=(273−45+72)−(0)= \left(\frac{27}{3} - 45 + 72\right) - (0)

=9−45+72= 9 - 45 + 72

=36= 36

infoNote

Fractions commonly appear in definite integrals because of the division in the primitive formula. Take care with common denominators and simplifying fractions.

Writing the integrand as separate fractions

chatImportant

Since there is no quotient rule for integration, when the integrand is a fraction with multiple terms in the numerator, we must split it into separate fractions.

lightbulbExample

Write as separate fractions then evaluate:

a) ∫123x4−2x2x2 dx\int_1^2 \frac{3x^4 - 2x^2}{x^2} \, dx

Solution:

Divide each term in the numerator by x2x^2:

3x4x2=3x2 and 2x2x2=2\frac{3x^4}{x^2} = 3x^2 \text{ and } \frac{2x^2}{x^2} = 2

∫123x4−2x2x2 dx=∫12(3x2−2) dx\int_1^2 \frac{3x^4 - 2x^2}{x^2} \, dx = \int_1^2 (3x^2 - 2) \, dx

=[x3−2x]12= [x^3 - 2x]_1^2

=(8−4)−(1−2)= (8 - 4) - (1 - 2)

=4−(−1)= 4 - (-1)

=5= 5

b) ∫−3−2x3−2x4x3 dx\int_{-3}^{-2} \frac{x^3 - 2x^4}{x^3} \, dx

Solution:

Divide both terms by x3x^3:

x3x3=1 and 2x4x3=2x\frac{x^3}{x^3} = 1 \text{ and } \frac{2x^4}{x^3} = 2x

∫−3−2x3−2x4x3 dx=∫−3−2(1−2x) dx\int_{-3}^{-2} \frac{x^3 - 2x^4}{x^3} \, dx = \int_{-3}^{-2} (1 - 2x) \, dx

=[x−x2]−3−2= [x - x^2]_{-3}^{-2}

=(−2−4)−(−3−9)= (-2 - 4) - (-3 - 9)

=−6−(−12)= -6 - (-12)

=6= 6

Working with negative indices

The fundamental theorem works equally well with negative powers. However, we need to carefully convert between negative powers and fractions.

lightbulbExample

Use negative indices to evaluate:

a) ∫15x−2 dx\int_1^5 x^{-2} \, dx

Solution:

Increase the index from −2-2 to −1-1, then divide by −1-1

∫15x−2 dx=[x−1−1]15\int_1^5 x^{-2} \, dx = \left[\frac{x^{-1}}{-1}\right]_1^5

=[−1x]15= \left[-\frac{1}{x}\right]_1^5

Rewrite x−1x^{-1} as 1x\frac{1}{x} before substituting

=−15−(−11)= -\frac{1}{5} - \left(-\frac{1}{1}\right)

=−15+1= -\frac{1}{5} + 1

=45= \frac{4}{5}

b) ∫121x4 dx\int_1^2 \frac{1}{x^4} \, dx

Solution:

First rewrite as a negative power: 1x4=x−4\frac{1}{x^4} = x^{-4}

∫121x4 dx=∫12x−4 dx\int_1^2 \frac{1}{x^4} \, dx = \int_1^2 x^{-4} \, dx

Increase the index to −3-3 and divide by −3-3

=[x−3−3]12= \left[\frac{x^{-3}}{-3}\right]_1^2

=[−13x3]12= \left[-\frac{1}{3x^3}\right]_1^2

Rewrite as a fraction before substituting

=−13(8)−(−13(1))= -\frac{1}{3(8)} - \left(-\frac{1}{3(1)}\right)

=−124+13= -\frac{1}{24} + \frac{1}{3}

=−124+824= -\frac{1}{24} + \frac{8}{24}

=724= \frac{7}{24}

Why n = -1 is special

chatImportant

The power rule for primitives does not work when n=−1n = -1 because it would require division by zero:

∫x−1 dx=∫1x dx=[x00]=nonsense!\int x^{-1} \, dx = \int \frac{1}{x} \, dx = \left[\frac{x^0}{0}\right] = \text{nonsense!}

This special case requires logarithmic functions, which you will study later.

Warning: do not integrate across an asymptote

chatImportant

This is extremely important: You cannot integrate across a point where the function is undefined (has an asymptote).

Incorrect example:

Consider this calculation: ∫−11x−4 dx\int_{-1}^{1} x^{-4} \, dx

Following the usual method would give:

∫−11x−4 dx=[x−3−3]−11\int_{-1}^{1} x^{-4} \, dx = \left[\frac{x^{-3}}{-3}\right]_{-1}^{1}

=−13−13= -\frac{1}{3} - \frac{1}{3}

=−23= -\frac{2}{3}

Why this is nonsense: The function y=x−4=1x4y = x^{-4} = \frac{1}{x^4} has a vertical asymptote at x=0x = 0. The function is not defined at x=0x = 0, so we cannot integrate from −1-1 to 11 because this interval includes x=0x = 0.

Additionally, since y=x−4y = x^{-4} is always positive, how could the integral (which represents area) be negative?

Always check: Before integrating, check whether there are any asymptotes or discontinuities between the limits of integration.

Remember!

bookmarkSummary

Key Points to Remember:

  • A primitive F(x)F(x) of a function f(x)f(x) satisfies F′(x)=f(x)F'(x) = f(x). Finding primitives reverses differentiation.

  • The fundamental theorem of calculus states that ∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a), where F(x)F(x) is any primitive of f(x)f(x). This connects the processes of differentiation and integration.

  • To evaluate a definite integral: Find a primitive, write it in square brackets with limits, substitute the upper limit, substitute the lower limit, then subtract.

  • Expand brackets first when products appear (no product rule for integration), and split fractions when the numerator has multiple terms (no quotient rule for integration).

  • With negative indices: Convert between negative powers and fractions carefully. The rule works for all n≠−1n \neq -1.

  • Never integrate across an asymptote where the function is undefined. Always check for discontinuities within the interval of integration.

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