Definite integrals as signed area

• Interpret a definite integral as net signed area across a fixed interval.
• Separate a graph at its axis crossings before combining geometric regions.
• Verify a geometric integral by evaluating an antiderivative at the bounds.

Build the idea

Worked example

1. Step 1

   Set the line height x minus one equal to zero. The graph crosses the horizontal axis at x equal to one, which separates negative and positive contributions.

   $$x-1=0 \quad\Longrightarrow\quad x=1$$

2. Step 2

   From zero to one, the triangle has base one and height one. Its geometric area is one half, but its location below the axis gives a signed contribution of negative one half.

   $$-\frac{1}{2}(1)(1)=-\frac{1}{2}$$

3. Step 3

   From one to four, the triangle has base three and height three. It lies above the axis, so its positive contribution is nine halves.

   $$\frac{1}{2}(3)(3)=\frac{9}{2}$$

4. Step 4

   Add the signed contributions rather than their magnitudes. The larger positive triangle exceeds the negative triangle by four square units.

   $$-\frac{1}{2}+\frac{9}{2}=4$$

5. Step 5

   An antiderivative is one half x squared minus x. Evaluating at four and zero reproduces the same net accumulation.

   $$\left[\frac{x^2}{2}-x\right]_{0}^{4}=4$$

• Adding both triangle magnitudes as positive and calling the result the definite integral.

  That calculation gives total geometric area, which would be five. The definite integral uses orientation, so the triangle below the axis contributes negative one half and the net is four.

• Using the interval endpoints as one triangle even though the graph crosses the axis inside the interval.

  Split at every axis crossing because the sign changes there. Treating the entire interval as one unsigned shape loses the distinction between positive and negative accumulation.

Find the signed area under y equals x minus two from x equal to zero to x equal to five, separating the regions at the axis crossing.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Read height with orientation

   Graph y equals x minus one from zero to four. The integral accumulates narrow strips of height times width. Strips above the horizontal axis have positive height, while strips below it have negative height. That orientation is why the final number measures net accumulation instead of adding every geometric region as positive.

   • The integral from zero to four of x minus one with respect to x.
2. Part 2

   Separate where the sign changes

   The line crosses the axis when x minus one equals zero, so split the interval at x equal to one. From zero to one, a triangle lies below the axis. From one to four, a larger triangle lies above it. Handling the regions separately keeps their signs visible.

   • x minus one equals zero, which implies x equals one.
3. Part 3

   Calculate the net area

   The first triangle has base one and height one, giving magnitude one half and signed value negative one half. The second has base three and height three, giving positive nine halves. Their signed sum is four. Adding magnitudes instead would answer a different question about total geometric area.

   • Negative one half times one times one equals negative one half.
   • One half times three times three equals nine halves.
   • Negative one half plus nine halves equals four.
4. Part 4

   Check with the Fundamental Theorem

   Use the antiderivative one half x squared minus x. Its value at four is four, and its value at zero is zero. Upper value minus lower value equals four, exactly matching the geometric result. Agreement between the two approaches helps catch a lost negative sign or an incorrect triangle dimension.

   • The quantity x squared over two minus x, evaluated from zero to four, equals four.
   • Negative one half plus nine halves equals four.

Scope and limitations

• This lesson uses a linear graph with triangular regions and does not derive Riemann sums or cover numerical approximation for functions without elementary antiderivatives.
• Signed area is a useful geometric interpretation for scalar graphs, but applications with units may represent displacement, accumulated change, or another quantity rather than literal area.