The Fundamental Theorem of Calculus

Expert reviewed • 04 March 2025 • 9 minute read

HSC Maths Advanced Syllabus

  • establish and use the formula∫xn.dx=1n+1xn+1+C\int x^n.dx=\frac{1}{n+1}x^{n+1}+C for n≠1n\ne 1
  • use the formula ∫abf(x)dx=F(b)−F(a)\int_a^bf(x)dx=F(b)-F(a), where F(x)F(x) is the anti-derivative of f(x)f(x), to calculate definite integrals
    • understand and use the Fundamental Theorem of Calculus,F′(x)=ddx[∫axf(t)dt]=f(x)F'(x)=\frac{d}{dx}[\int_a^xf(t)dt]=f(x) and illustrate its proof geometrically

Note:

Video coming soon!

How are Primitives Related to Integration?

As we know primitives are the anti-derivative of a function, which can be found by using integration. As such, there is a simple, general formula that can be used to find the primitive of a function of the form: xnx^n, where n≠1n\ne1.

∫xn.dx=1n+1xn+1+C\int x^n.dx=\frac{1}{n+1}x^{n+1}+C

Where CC is equal to some constant. We will touch on the concept of the constant CC while integrating, in a later chapter.

What is the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus connects the two central operations of calculus: differentiation and integration. It essentially states that a definite integral can be calculated. This is most commonly completed by determining the primitive of a function's derivative, using the formula above, and substituting limits into the equation.

There are two main parts to this theorem:

  • If a function f(x)f(x) is governed by the limits [a,b][a,b], and it's anti-derivative is F(x)F(x), then we can apply the following formula to determine the definite integral.
∫abf(x)dx=F(b)−F(a)\int_a^bf(x)dx=F(b)-F(a)

This is often used to calculate the area under the curve, created by a function, as your evaluated answer will be a number.

  • If a function f(x)f(x) is has limits with an open interval, for example [a,x][a,x] where aa is a constant, and it's antiderivative is F(x)F(x), it is defined by the formula:
F(x)=∫axf(t)dtF(x)=\int_a^xf(t)dt

This part is often referred to as the Second Fundamental Theorem of Calculus. It tells us that if you construct a function F(x)F(x) by integrating another function f(x)f(x) from one fixed point aa, then the integral of f(x)f(x) is just the original function. This formula is generally noted as the signed area function, as it is commonly used to determine the areas of a graph that are located in both the negative and positive axis.

Practice Question 1

Evaluate the following expression: ∫02x(x+1)dx\int_0^{2} x(x+1)dx

Solution

As explained above, we must use the formula related to the first part of the fundamental theorem, as we are dealing with an integral with limits in the form [a,b][a,b].

However, we must first simplify the integral, so we can apply the formula to find a functions primitive.

∫02x(x+1)dx=∫02(x2+x)dx\int_0^{2}x(x+1)dx=\int_0^{2}(x^2+x)dx

Now we can integrate:

∫02(x2+x)dx=[(12+1x2+1)+(11+1x1+1)]02=[x33+x22]02\int_0^{2}(x^2+x)dx =[(\frac{1}{2+1}x^{2+1})+(\frac{1}{1+1}x^{1+1})]^{2}_{0}\\=[\frac{x^3}{3}+\frac{x^2}{2}]^{2}_0

Using the fundamental theorem formula: ∫abf(x)dx=F(b)−F(a)\int_a^bf(x)dx=F(b)-F(a) we can now evaluate the integral:

We do this by substituting b=2b=2 and a=0a=0 into the equation we have found.

[x33+x22]02=(233+222)−(033+022)=(83+2)−0=143[\frac{x^3}{3}+\frac{x^2}{2}]^{2}_0=(\frac{2^3}{3}+\frac{2^2}{2})-(\frac{0^3}{3}+\frac{0^2}{2})\\=(\frac{8}{3}+2)-0\\=\frac{14}{3}

Practice Question 2

Integrate the following: f(x)=∫0x3x2.dxf(x)=\int_0^x3x^2.dx

Solution

As explained above, we must use the signed area formula related to the second part of the fundamental theorem, as we are dealing with an integral with limits in the form [a,x][a,x].

The integral has already been simplified as much as it can.

Now we must integrate by applying the formula to find the primitive:

∫0x3x2dx=[12+13x2+1]0x=[3x33]0x=[x3]0x\int_0^{x}3x^2dx =[\frac{1}{2+1}3x^{2+1}]^{x}_{0}\\= [\frac{3x^3}{3}]_0^x\\=[x^3]_0^x

Using the fundamental theorem formulas: F(x)=∫axf(t)dtF(x)=\int_a^xf(t)dt and ∫abf(x)dx=F(b)−F(a)\int_a^bf(x)dx=F(b)-F(a)we can now determine the original function:

F(x)=(x3)−(03)=x3F(x)=(x^3)-(0^3)\\=x^3

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Introduction to Integration

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The Definite Integral

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