INTEGRATION 03

Definite Integrals

A definite integral measures the accumulated effect of a function over a specified interval.

The main idea

The notation

\[ \boxed{ \int_a^b f(x)\,dx } \]

represents the accumulation of \(f(x)\) from \(x=a\) to \(x=b\).

1. Lower and upper limits

\[ \int_a^b f(x)\,dx, \]

\(a\) is the lower limit and \(b\) is the upper limit.

They specify the interval over which the accumulation is measured.

2. Definite versus indefinite integrals

An indefinite integral gives a family of antiderivatives:

\[ \int f(x)\,dx=F(x)+C. \]

A definite integral gives a single numerical value:

\[ \int_a^b f(x)\,dx. \]

So these are related, but they are not the same object.

3. Area interpretation

\[ f(x)\ge0 \]

on the interval \([a,b]\), then

\[ \int_a^b f(x)\,dx \]

gives the area between the graph, the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\).

4. Example

Consider

\[ \int_0^2 x\,dx. \]

An antiderivative of \(x\) is

\[ \frac{x^2}{2}. \]

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

Evaluate at the upper limit:

\[ \frac{2^2}{2}=2. \]

Evaluate at the lower limit:

\[ \frac{0^2}{2}=0. \]

Therefore,

\[ \boxed{ \int_0^2 x\,dx=2 } \]

5. The evaluation notation

The notation

\[ \left[F(x)\right]_a^b \]

means

\[ \boxed{ F(b)-F(a) } \]

So we evaluate the antiderivative at the upper limit and subtract its value at the lower limit.

6. Why there is no \(+C\)

Suppose an antiderivative is

\[ F(x)+C. \]

Then

\[ [F(b)+C]-[F(a)+C]. \]

The constants cancel:

\[ F(b)-F(a). \]

That is why a definite integral does not need a \(+C\).

7. Signed area

A definite integral gives signed accumulation, not always ordinary geometric area.

\[ f(x)>0, \]

the contribution is positive.

\[ f(x)<0, \]

the contribution is negative.

So regions below the \(x\)-axis subtract from the integral.

8. Example with negative contribution

Consider

\[ \int_{-1}^{1}x\,dx. \]

An antiderivative is

\[ \frac{x^2}{2}. \]

Therefore,

\[ \int_{-1}^{1}x\,dx = \left[ \frac{x^2}{2} \right]_{-1}^{1}. \]

\[ = \frac12-\frac12 = 0. \]

The positive and negative contributions cancel.

9. Integral versus total geometric area

If a function crosses the \(x\)-axis, the definite integral may be smaller than the total geometric area.

To find total geometric area, regions below the axis must be counted positively.

This may require splitting the interval where the function changes sign.

10. Reversing the limits

If the limits are reversed,

\[ \boxed{ \int_b^a f(x)\,dx = -\int_a^b f(x)\,dx } \]

Changing the direction of accumulation changes the sign.

11. Same upper and lower limit

If the interval has zero width,

\[ a=b, \]

then

\[ \boxed{ \int_a^a f(x)\,dx=0 } \]

because there is no interval over which to accumulate.

12. Splitting an interval

\[ a then

\[ \boxed{ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx } \]

Accumulation over the whole interval equals the sum of accumulation over its parts.

13. Example with a polynomial

Evaluate

\[ \int_1^3(2x+1)\,dx. \]

An antiderivative is

\[ x^2+x. \]

Therefore,

\[ \int_1^3(2x+1)\,dx = [x^2+x]_1^3. \]

\[ =(9+3)-(1+1) \] \[ =12-2 \] \[ =10. \]

Hence,

\[ \boxed{ \int_1^3(2x+1)\,dx=10 } \]

14. Units of a definite integral

If \(f(x)\) has units and \(x\) has units, the integral combines them.

For example, if

\[ v(t) \]

is measured in metres per second and \(dt\) represents seconds, then

\[ v(t)\,dt \]

has units of metres.

Therefore,

\[ \int_a^b v(t)\,dt \]

has units of displacement.

15. Rate to total change

Suppose

\[ \frac{dy}{dx}=f(x). \]

Then the accumulated change in \(y\) from \(x=a\) to \(x=b\) is

\[ \boxed{ y(b)-y(a) = \int_a^b f(x)\,dx } \]

This is one of the central meanings of a definite integral.

16. The deeper connection

We began with integration as accumulation. We then learned that antiderivatives reverse differentiation.

A remarkable result connects these ideas:

\[ \boxed{ \int_a^b f(x)\,dx = F(b)-F(a), \qquad F'(x)=f(x) } \]

This is the Fundamental Theorem of Calculus.