Search topics

Search

The area function and the fundamental theorem of calculus

Start a set of 8
Step 2 of 5

The fundamental theorem of calculus

Let f be continuous on an interval and let F be an anti-derivative of f there. With a fixed lower terminal a, the area function accumulates signed area along that interval. At an interior input, its derivative is the integrand evaluated there. A definite integral is the upper primitive value minus the lower primitive value

\[A(x)=\int_a^x f(t)\,dt\Rightarrow A'(x)=f(x),\quad\int_a^b f(x)\,dx=F(b)-F(a)\]
  1. 1Substitute the upper terminal Replace the dummy variable by the variable upper terminal when differentiating an area function \(\displaystyle \frac{d}{dx}\int_1^x\left(2t+3\right)\,dt=2x+3\)
  2. 2Treat the dummy variable correctly The letter inside the integral can change because it is only a placeholder \(\displaystyle \int_1^x\left(2t+3\right)\,dt=\int_1^x\left(2u+3\right)\,du\)
  3. 3Subtract lower from upper Evaluate an anti-derivative at both terminals and subtract the lower value \(\displaystyle \int_1^3 2x\,dx=F(3)-F(1)=8\)
  4. 4Cancel the arbitrary constant The same constant occurs at both terminals and cancels in a definite integral \(\displaystyle \left[x^2+C\right]_1^3=\left(9+C\right)-\left(1+C\right)=8\)
Year 12, VCE Units 3 & 4, Chapter 8, 4 difficulty levels, No calculatorBuild a worksheet

The fundamental theorem of calculus

Let f be continuous on an interval and let F be an anti-derivative of f there, so F′ = f. With a fixed lower terminal a, the area function accumulates signed area along that interval. At an interior input x, its derivative is f(x). A definite integral is the upper primitive value minus the lower primitive value

\[A(x)=\int_a^x f(t)\,dt\Rightarrow A'(x)=f(x),\quad\int_a^b f(x)\,dx=F(b)-F(a)\]
  1. Substitute the upper terminal Replace the dummy variable by the variable upper terminal when differentiating an area function \(\displaystyle \frac{d}{dx}\int_1^x\left(2t+3\right)\,dt=2x+3\)
  2. Treat the dummy variable correctly The letter inside the integral can change because it is only a placeholder \(\displaystyle \int_1^x\left(2t+3\right)\,dt=\int_1^x\left(2u+3\right)\,du\)
  3. Subtract lower from upper Evaluate an anti-derivative at both terminals and subtract the lower value \(\displaystyle \int_1^3 2x\,dx=F(3)-F(1)=8\)
  4. Cancel the arbitrary constant The same constant occurs at both terminals and cancels in a definite integral \(\displaystyle \left[x^2+C\right]_1^3=\left(9+C\right)-\left(1+C\right)=8\)

Worked example

For the area function below, find A'(x) and A(3)

\[A(x)=\int_1^x\left(2t+3\right)\,dt\]
  1. \(\displaystyle A'(x)=2x+3\)The fundamental theorem replaces the dummy variable t by the variable upper terminal x
  2. \(\displaystyle F(x)=x^2+3x\)Integrating each term gives this anti-derivative
  3. \(\displaystyle A(3)=F(3)-F(1)=(3^2+3(3))-(1^2+3(1))=18-4=14\)Use the upper primitive value minus the lower primitive value

Common mistakes

Not substituting the upper terminal
✕\(\displaystyle \frac{d}{dx}\int_1^x\left(2t+3\right)\,dt=2t+3\)✓\(\displaystyle \frac{d}{dx}\int_1^x\left(2t+3\right)\,dt=2x+3\)
The dummy variable t is replaced by the variable upper terminal x in the derivative
Using the area function as its derivative
✕\(\displaystyle A'(x)=A(x)\)✓\(\displaystyle A'(x)=2x+3\)
The derivative of the accumulated area is the integrand evaluated at the upper terminal
Keeping the arbitrary constant
✕\(\displaystyle \int_1^3 2x\,dx=\left[x^2+C\right]_1^3=8+C\)✓\(\displaystyle \int_1^3 2x\,dx=\left[x^2\right]_1^3=8\)
The same arbitrary constant appears at both terminals and cancels

Difficulty levels

introDifferentiate a simple area function with a polynomial integrand
standardUse the upper-minus-lower rule with a multi-term integrand
challengeDistinguish the two theorem statements or use an exponential integrand
examChoose the correct theorem line and calculate an exact value

Questions about this topic

Why can the dummy variable be changed?+

The variable inside the integral is a placeholder. Changing t to u does not change the accumulated function when the limits and integrand are changed consistently

Why is the lower terminal subtracted from the upper terminal?+

The definite-integral rule is upper primitive value minus lower primitive value, F(b)−F(a)

What does an area function measure?+

It measures the signed accumulated area from a fixed lower terminal to a variable upper terminal

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 8: Integration

In this chapter

Approximate area by an endpoint or trapezium ruleEvaluate and transform a definite integralElementary primitives and exact definite integralsFind complete families of polynomial anti-derivativesThe area function and the fundamental theorem of calculusSigned integrals, geometric area and motionFind the area between two curvesAverage value of a functionTotal change and kinematics from a rateDetermine a polynomial from its derivative and a boundary conditionRecover motion from acceleration