Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) establishes the critical link between differentiation and integration, showing they are inverse processes. It provides the standard method for evaluating definite integrals, which is essential for calculating quantities like the area under a curve.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The theorem comes in two main parts, both connecting derivatives and integrals.
- Part 1: The definite integral of a function f(x) from a to b is the net change in its antiderivative, F(x), between those points.
- Part 2: The derivative of an integral with a fixed lower limit and a variable upper limit 'x' is simply the original function evaluated at 'x'.
- A key consequence is that swapping the limits of a definite integral negates its value: ∫a^b f(x) dx = -∫b^a f(x) dx.
- You can split an integral into parts: ∫a^c f(x) dx + ∫c^b f(x) dx = ∫a^b f(x) dx for any a, b, and c (provided the function is defined).
Diagram
Formulae
∫a^b f(x) dx = F(b) - F(a), where F'(x) = f(x) Use this to calculate the exact numerical value of a definite integral. First find the antiderivative F(x), then evaluate it at the upper limit (b) and subtract its value at the lower limit (a).
d/dx [∫a^x f(t) dt] = f(x) Use this when asked to differentiate an integral where the upper limit is the variable 'x'. The process of integration is undone by differentiation, leaving just the original function.
Definitions
- Antiderivative
- A function F(x) is an antiderivative of f(x) if F'(x) = f(x). For example, x3 is an antiderivative of 3x2.
- Definite Integral
- An integral with specified upper and lower limits, written as ∫a^b f(x) dx. It evaluates to a single numerical value, representing the signed area between the curve, the x-axis, and the lines x=a and x=b.
- Fundamental Theorem of Calculus (FTC)
- The core theorem connecting differentiation and integration, allowing for the precise evaluation of definite integrals using antiderivatives.
Worked example
A function is defined as G(x) = d/dx [∫1^x (t2 + 1)3 dt]. What is the value of G(2)?
- 1
First, evaluate the inner expression using the Fundamental Theorem of Calculus, Part 2.
- 2
Identify the integrand f(t) = (t2 + 1)3.
- 3
According to the theorem, d/dx [∫a^x f(t) dt] = f(x).
- 4 So, G(x) = (x2 + 1)3
- 5
Now, substitute x = 2 into the expression for G(x).
- 6 G(2) = (22 + 1)3 = (4 + 1)3 = 53
- 7
Calculate the final value:
53 = 125
Answer: 125
Common mistakes
- ×When calculating ∫a^b f(x) dx, a very common error is mixing up the order and calculating F(a) - F(b) instead of F(b) - F(a). Always do (upper limit) minus (lower limit).
- ×Forgetting that the constant of integration, '+ C', is not needed for definite integrals because it cancels out during the subtraction: (F(b) + C) - (F(a) + C) = F(b) - F(a).
- ×Misapplying the second part of the theorem by trying to integrate first. The rule d/dx [∫a^x f(t) dt] = f(x) is a direct shortcut that avoids any integration.
No-calculator tips
- ✓When calculating F(b) - F(a), work out F(b) and F(a) as two separate, clear calculations on your rough paper before performing the subtraction. This minimises the chance of arithmetic or sign errors.
- ✓If the antiderivative F(x) involves multiple terms or fractions, group the F(b) and F(a) terms in brackets to keep track of signs, for example: (term1(b) + term2(b)) - (term1(a) + term2(a)).