The fundamental theorem of calculus
Understand the fundamental theorem of calculus in Year 12 Specialist Mathematics for Queensland (QCAA) — the result that ties differentiation and integration together as inverse processes. One form differentiates an accumulation function; the other evaluates a definite integral as \(F(b)-F(a)\).
You will learn to differentiate integrals with variable and function upper limits using the chain rule, evaluate definite integrals from any antiderivative, and read an accumulation function as an area — the foundation for areas, volumes and further applications of integral calculus in Unit 4.
Theory
The fundamental theorem of calculus is the bridge between the two halves of Year 12 Specialist Mathematics calculus (QCAA, Queensland): it says differentiation and integration are inverse processes. One form differentiates an accumulation function, \(\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)\); the other evaluates a definite integral, \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\).
An accumulation function \(A(x)=\displaystyle\int_a^x f(t)\,dt\) measures the signed area under \(y=f(t)\) from a fixed left edge \(t=a\) to a moving right edge \(t=x\). The upper limit is the variable \(x\); the dummy variable \(t\) is just a placeholder inside the integral.
The fundamental theorem of calculus (derivative form) says that differentiating this accumulation function gives back the integrand: \(\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)\). In words, the rate at which the area grows as the right edge slides is exactly the height \(f(x)\) of the curve at that edge. The fixed lower limit \(a\) does not affect the derivative.
When the upper limit is a function \(g(x)\) rather than \(x\) itself, the chain rule joins in: \(\dfrac{d}{dx}\displaystyle\int_a^{g(x)} f(t)\,dt=f\big(g(x)\big)\,g'(x)\). If the variable sits in the lower limit, reverse the limits first, which flips the sign.
The fundamental theorem of calculus (evaluation form) turns the theorem around: to evaluate \(\displaystyle\int_a^b f(x)\,dx\), find any antiderivative \(F\) of \(f\) and compute \(F(b)-F(a)\). Because integrating then differentiating (and differentiating then integrating) return the original function, differentiation and integration are inverse processes.
Fundamental theorem of calculus — derivative form (variable upper limit):
With a function as the upper limit, the chain rule joins in:
If both limits are functions of \(x\) (Leibniz rule):
Fundamental theorem of calculus — evaluation form, for any antiderivative \(F\) with \(F'=f\):
Using the fundamental theorem of calculus
- Spot the form: a derivative of an integral (a \(\frac{d}{dx}\) outside) uses the derivative form; a plain definite integral \(\int_a^b\) uses the evaluation form.
- Derivative form: read off the integrand at the upper limit — if the upper limit is just \(x\), the answer is \(f(x)\); if it is a function \(g(x)\), multiply by \(g'(x)\).
- Check the limits: if the variable is in the lower limit, reverse the limits first (this flips the sign); if both limits vary, subtract the lower contribution.
- Evaluation form: find any antiderivative \(F\), then substitute to get \(F(b)-F(a)\) — no constant of integration is needed.
The upper limit is just \(x\), so the fundamental theorem returns the integrand with \(t\) replaced by \(x\) — no antiderivative needed:
| \(\dfrac{d}{dx}\int_a^x f(t)\,dt\) | \(=\) | \(f(x)\) |
| \(f(t)\) | \(=\) | \(\sqrt{1+t^3}\) |
| \(\dfrac{d}{dx}\int_{3}^{x}\sqrt{1+t^3}\,dt\) | \(=\) | \(\sqrt{1+x^3}\) |
\(\dfrac{d}{dx}\displaystyle\int_{3}^{x}\sqrt{1+t^3}\,dt=\sqrt{1+x^3}\).
The upper limit is a function \(g(x)=x^2\), so combine the theorem with the chain rule: \(\frac{d}{dx}\int_a^{g(x)} f(t)\,dt=f(g(x))\,g'(x)\).
| \(g(x)\) | \(=\) | \(x^2\) |
| \(g'(x)\) | \(=\) | \(2x\) |
Evaluate the integrand at \(g(x)\) and multiply by \(g'(x)\):
| \(\dfrac{d}{dx}\int_{0}^{x^2}\cos t\,dt\) | \(=\) | \(\cos\!\big(x^2\big)\cdot 2x\) |
| \(=\) | \(2x\cos\!\big(x^2\big)\) |
\(\dfrac{d}{dx}\displaystyle\int_{0}^{x^2}\cos t\,dt=2x\cos\!\big(x^2\big)\).
Find an antiderivative \(F\) term by term (no constant needed for a definite integral):
| \(F(x)\) | \(=\) | \(\int\big(6x^2-2x\big)\,dx\) |
| \(=\) | \(2x^3-x^2\) |
Substitute the limits and subtract, \(F(2)-F(1)\):
| \(\int_{1}^{2}\big(6x^2-2x\big)\,dx\) | \(=\) | \(\Big[2x^3-x^2\Big]_{1}^{2}\) |
| \(=\) | \(\big(16-4\big)-\big(2-1\big)\) | |
| \(=\) | \(12-1\) | |
| \(=\) | \(11\) |
\(\displaystyle\int_{1}^{2}\big(6x^2-2x\big)\,dx=11\).
Evaluate the integral to write \(A(x)\) as an explicit function:
| \(A(x)\) | \(=\) | \(\left[\dfrac{t^2}{2}-2t\right]_{0}^{x}\) |
| \(=\) | \(\dfrac{x^2}{2}-2x\) |
Differentiate; by the fundamental theorem this returns the integrand:
| \(A'(x)\) | \(=\) | \(\dfrac{d}{dx}\int_{0}^{x}(t-2)\,dt\) |
| \(=\) | \(x-2\) |
The minimum occurs where \(A'(x)=0\):
| \(x-2\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(2\) |
\(A(x)=\dfrac{x^2}{2}-2x\), \(A'(x)=x-2\), and \(A\) is least at \(x=2\).
Common pitfalls
Frequently asked questions
What is the fundamental theorem of calculus?
It links differentiation and integration as inverse processes. The derivative form is \(\frac{d}{dx}\int_a^x f(t)\,dt=f(x)\); the evaluation form is \(\int_a^b f(x)\,dx=F(b)-F(a)\), where \(F\) is any antiderivative of \(f\).
How do you differentiate an integral with a variable upper limit?
By the fundamental theorem, \(\frac{d}{dx}\int_a^x f(t)\,dt=f(x)\): just replace the dummy variable \(t\) with \(x\) in the integrand. The fixed lower limit \(a\) plays no part.
What happens when the upper limit is a function like x squared?
Use the chain rule: \(\frac{d}{dx}\int_a^{g(x)} f(t)\,dt=f(g(x))\,g'(x)\). For example \(\frac{d}{dx}\int_0^{x^2}\cos t\,dt=2x\cos(x^2)\).
Do I need the constant of integration when evaluating a definite integral?
No. Any antiderivative works, so the \(+c\) cancels in \(F(b)-F(a)\). You only add \(+c\) for an indefinite integral.
What if the variable is in the lower limit of the integral?
Reverse the limits first, which flips the sign: \(\frac{d}{dx}\int_x^b f(t)\,dt=-\frac{d}{dx}\int_b^x f(t)\,dt=-f(x)\).
What is an accumulation function?
A function defined by an integral with a variable upper limit, \(A(x)=\int_a^x f(t)\,dt\). It measures the signed area swept out under \(f\) up to \(x\), and its derivative is \(A'(x)=f(x)\).