Knowledge Science · The definite integral

1. Evaluate at the endpoints

1 / 4

Example graph

\int_0^b x^2\,dx=\left[\frac{x^3}{3}\right]_0^b

For x², a primitive is x³/3. Subtract its value at the lower endpoint from its value at the upper endpoint. The constant cancels.

F(b): 0.333333
F(0): 0
∫: 0.333333

Shading shows the region between the bounds. With bounds in increasing order, regions below the axis contribute negatively; reversing the bounds reverses the integral’s sign. Geometric area stays nonnegative.

2. Reverse the interval

2 / 4

Example graph

\int_1^b x\,dx=\frac{b^2-1}{2}

The integral from 1 to b of x is (b²−1)/2. When b<1, the orientation reverses and the integral is negative even though x is positive on the shaded region.

b: 2
∫₁ᵇ x dx: 1.5

Shading shows the region between the bounds. With bounds in increasing order, regions below the axis contribute negatively; reversing the bounds reverses the integral’s sign. Geometric area stays nonnegative.

3. Displacement and distance

3 / 4

Example graph

v(t)=t-1,\quad\Delta s=\int_0^b v(t)\,dt

With velocity v(t)=t−1 on [0,2], negative and positive contributions cancel. Displacement is ∫v; total distance is ∫|v| and cannot cancel.

Δs: -0.5
distance: 0.5

Shading shows the region between the bounds. With bounds in increasing order, regions below the axis contribute negatively; reversing the bounds reverses the integral’s sign. Geometric area stays nonnegative.

4. Average function value

4 / 4

Example graph

f_{avg}=\frac1b\int_0^b x^2\,dx=\frac{b^2}{3}

The average height on [0,b], b>0, is (1/b)∫₀ᵇ x² dx=b²/3. This is the height of a rectangle with the same base and area as the region.

∫: 0.333333
average: 0.333333

Shading shows the region between the bounds. With bounds in increasing order, regions below the axis contribute negatively; reversing the bounds reverses the integral’s sign. Geometric area stays nonnegative.