The Fundamental Theorem of Calculus and the Definite Integral (VCE SSCE Mathematical Methods): Revision Notes

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The Fundamental Theorem of Calculus and the Definite Integral

Introduction to definite and indefinite integrals

In previous sections, you learned about indefinite integrals. These are called "indefinite" because they include an arbitrary constant cc in their solution. For example, when we integrate 3x23x^2, we write:

∫3x2 dx=x3+c\int 3x^2 \, dx = x^3 + c

More generally, we express this as:

∫f(x) dx=F(x)+c\int f(x) \, dx = F(x) + c

where F(x)F(x) is an antiderivative of f(x)f(x).

infoNote

The constant cc appears in indefinite integrals because differentiation of a constant equals zero, meaning there are infinitely many antiderivatives that differ only by a constant value.

Now we'll explore the definite integral and discover its powerful connection to the indefinite integral through the Fundamental Theorem of Calculus.

Signed area

When working with definite integrals, we need to understand the concept of signed area. This is crucial because the area under a curve can be above or below the xx-axis.

Regions above and below the x-axis

chatImportant

Key principle:

  • Regions above the x-axis have positive signed area
  • Regions below the x-axis have negative signed area

Example: the line y = x + 1

Consider the graph of y=x+1y = x + 1. Looking at the shaded regions:

A1=12×3×3=412A_1 = \frac{1}{2} \times 3 \times 3 = 4\frac{1}{2}

(triangle above the xx-axis)

A2=12×1×1=12A_2 = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}

(triangle below the xx-axis)

The total area is: A1+A2=5A_1 + A_2 = 5

The signed area is: A1−A2=4A_1 - A_2 = 4

General case with periodic functions

For a curve with multiple regions above and below the xx-axis (such as a sine wave), the calculations work as follows:

Total area: A1+A2+A3+A4A_1 + A_2 + A_3 + A_4

Signed area: A1−A2+A3−A4A_1 - A_2 + A_3 - A_4

Formal definition

infoNote

For any continuous function ff on an interval [a,b][a, b], the definite integral ∫abf(x) dx\int_a^b f(x) \, dx gives the signed area enclosed by the graph of y=f(x)y = f(x) between x=ax = a and x=bx = b.

This means the definite integral accounts for whether regions are above or below the xx-axis, making some contributions positive and others negative.

The Fundamental Theorem of Calculus

This theorem creates a remarkable bridge between antiderivatives and areas under curves. It tells us that we can calculate definite integrals by using antiderivatives.

Statement of the theorem

chatImportant

Fundamental Theorem of Calculus:

If ff is a continuous function on an interval [a,b][a, b], then:

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

where GG is any antiderivative of ff.

Notation for evaluation

To make our working clearer, we use the notation:

G(b)−G(a)=[G(x)]abG(b) - G(a) = \left[G(x)\right]_a^b

This notation means: "evaluate G(x)G(x) at the upper limit bb, then subtract the value of G(x)G(x) at the lower limit aa."

Evaluating definite integrals

Let's see how the Fundamental Theorem works in practice through several worked examples.

Worked example: basic polynomial

lightbulbExample

Evaluate: ∫12x dx\int_1^2 x \, dx

Solution:

First, find the antiderivative of xx:

∫x dx=12x2+c\int x \, dx = \frac{1}{2}x^2 + c

Now apply the Fundamental Theorem:

∫12x dx=[x22]12\int_1^2 x \, dx = \left[\frac{x^2}{2}\right]_1^2
chatImportant

The arbitrary constant cc cancels out when we subtract G(a)G(a) from G(b)G(b). Because of this, we can ignore the constant when evaluating definite integrals.

Worked example: power functions

lightbulbExample

Evaluate:

a) ∫23x2 dx\int_2^3 x^2 \, dx

b) ∫32x2 dx\int_3^2 x^2 \, dx

c) ∫01(x12+x32)dx\int_0^1 \left(x^{\frac{1}{2}} + x^{\frac{3}{2}}\right) dx

Solution:

a)

∫23x2 dx=[x33]23\int_2^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_2^3

b)

∫32x2 dx=[x33]32\int_3^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_3^2

Notice that this is the negative of part (a). This illustrates the property that reversing the limits changes the sign of the integral.

c)

∫01(x12+x32)dx=[23x32+25x52]01\int_0^1 \left(x^{\frac{1}{2}} + x^{\frac{3}{2}}\right) dx = \left[\frac{2}{3}x^{\frac{3}{2}} + \frac{2}{5}x^{\frac{5}{2}}\right]_0^1

Worked example: exponential functions

lightbulbExample

Evaluate:

a) ∫012e−2x dx\int_0^1 2e^{-2x} \, dx

b) ∫04(e2x+1) dx\int_0^4 (e^{2x} + 1) \, dx

c) ∫14(2x12+ex2)dx\int_1^4 \left(2x^{\frac{1}{2}} + e^{\frac{x}{2}}\right) dx

Solution:

a)

∫012e−2x dx=[2−2e−2x]01\int_0^1 2e^{-2x} \, dx = \left[\frac{2}{-2}e^{-2x}\right]_0^1

b)

∫04(e2x+1) dx=[12e2x+x]04\int_0^4 (e^{2x} + 1) \, dx = \left[\frac{1}{2}e^{2x} + x\right]_0^4

c)

∫14(2x12+ex2)dx=[43x32+2ex2]14\int_1^4 \left(2x^{\frac{1}{2}} + e^{\frac{x}{2}}\right) dx = \left[\frac{4}{3}x^{\frac{3}{2}} + 2e^{\frac{x}{2}}\right]_1^4

Worked example: logarithmic functions

lightbulbExample

Evaluate:

a) ∫681x−5 dx\int_6^8 \frac{1}{x - 5} \, dx

b) ∫4512x−5 dx\int_4^5 \frac{1}{2x - 5} \, dx

Solution:

a)

∫681x−5 dx=[log⁡e(x−5)]68\int_6^8 \frac{1}{x - 5} \, dx = \left[\log_e(x - 5)\right]_6^8

b)

∫4512x−5 dx=12[log⁡e(2x−5)]45\int_4^5 \frac{1}{2x - 5} \, dx = \frac{1}{2}\left[\log_e(2x - 5)\right]_4^5

Properties of the definite integral

Understanding these properties will help you manipulate and simplify definite integrals efficiently:

infoNote

Properties of Definite Integrals:

1. Additive property over intervals:

∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx

This means you can split an integral at any point cc between aa and bb.

2. Zero interval:

∫aaf(x) dx=0\int_a^a f(x) \, dx = 0

An integral over an interval with no width equals zero.

3. Constant multiple:

∫abkf(x) dx=k∫abf(x) dx\int_a^b k f(x) \, dx = k \int_a^b f(x) \, dx

You can factor out constants from integrals.

4. Sum and difference:

∫ab[f(x)±g(x)] dx=∫abf(x) dx±∫abg(x) dx\int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx

You can split the integral of a sum or difference into separate integrals.

5. Reversal of limits:

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

Swapping the upper and lower limits changes the sign of the integral.

Remember!

bookmarkSummary

Key Points to Remember:

  • The definite integral ∫abf(x) dx\int_a^b f(x) \, dx represents the signed area between the curve y=f(x)y = f(x) and the xx-axis from x=ax = a to x=bx = b.
  • Areas above the xx-axis contribute positively; areas below contribute negatively.
  • The Fundamental Theorem of Calculus states: ∫abf(x) dx=[G(x)]ab=G(b)−G(a)\int_a^b f(x) \, dx = [G(x)]_a^b = G(b) - G(a), where GG is any antiderivative of ff.
  • When evaluating definite integrals, the arbitrary constant cc always cancels out, so we don't need to include it.
  • Remember the notation: "top minus bottom" – evaluate at the upper limit and subtract the value at the lower limit.

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