The Definite Integral

The indefinite integral is a function; the definite integral is a number got by putting values into that function.

Let

F(x) = $$\int$$f (x) dx

So F is a function with the property that F' = f.

Definition 6..8   The definite integral of f from a to b is denoted by $\int_{a}^{b}$f (x) dx and is defined to be F(b) - F(a).

Definition 6..9   a and b are called the limits of integration.

Notation: $$\left[\vphantom{F(x)}\right.$$F(x)$$\left.\vphantom{F(x)}\right]_{a}^{b}$$ is used to mean F(b) - F(a).

Example 6..10  

$$\int_{1}^{2}$$x dx = $$\left[\vphantom{\frac{x^2}{2}}\right.$$$${\frac{{x^2}}{{2}}}$$$$\left.\vphantom{\frac{x^2}{2}}\right]_{1}^{2}$$ = 2 - $${\frac{{1}}{{2}}}$$ = $${\frac{{3}}{{2}}}$$

Remark 6..11   Note that the constant of integration is of no importance when you are calculating a definite integral. Include it and you get

(F(b) + c) - (F(a) + c) = F(b) - F(a)

-- the same answer as you would have got by ignoring the constant in the first place.

Remark 6..12   The definite integral is the one that is used to calculate areas, volumes, lengths, moments of inertia and a host of other things.

Properties of the Definite Integral

$$\int_{b}^{a} \int_{a}^{b}$$ (1)

So reversing the limits of integration multiplies the answer by -1.

Proof. Let F(x) be an indefinite integral for f (x). Then

$$\int_{b}^{a}$$ = F(a) - F(b) = - (F(b) - F(a))

$$\int_{a}^{b}$$

$$\int_{a}^{b} \int_{b}^{c} \int_{a}^{c}$$ (2)

Proof. F(x) as before. Then

$$\int_{a}^{b} \int_{b}^{c} \left(\vphantom{F(b)-F(a)}\right. \left.\vphantom{F(b)-F(a)}\right) \left(\vphantom{F(c)-F(b)}\right. \left.\vphantom{F(c)-F(b)}\right)$$ = F(c) - F(a)

$$\int_{a}^{c}$$

Using the Definite Integral to Calculate Areas

This is slightly more complicated than one might hope, but not by much. The problem is caused by the fact that definite integrals can be either positive or negative, while areas are always positive.

Case 1: The curve is entirely above the x-axis

This is complication free.

Choose a reference point (r, 0) on the x-axis, and let F(x) be the area under the section of the curve between (r, f (r)) and (x, f (x)).

As we saw earlier, F'(x) = f (x), and so F(x) = $\int$f (x) dx.

Now consider the area under the curve between x = a and x = b.

F(b) is the area between x = r and x = b; F(a) is that between x = r and x = a.

The cross-hatched area is the difference, i.e. F(b) - F(a). Therefore the area sought is $\int_{a}^{b}$f (x) dx.

Example 6..13   Find the area under the curve y = $\sqrt{x}$ between x = 0 and x = 4.

Solution The curve does not go below the x-axis and so

$$\int_{0}^{4} \sqrt{x}$$ Area 

$$\left[\vphantom{\frac{2}{3}x^{3/2}}\right. {\frac{{2}}{{3}}} \left.\vphantom{\frac{2}{3}x^{3/2}}\right]_{0}^{4}$$

$${\frac{{16}}{{3}}}$$

Case 2: The curve is entirely below the x-axis

This time y = f (x) is negative, and so the height of the rectangle we looked at when considering the difference between F(x₀ + h) and F(x₀) is not f (x₀) but - f (x₀) (since heights can't be negative). So F'(x) = - f (x), and so for areas under the x-axis

Area  = - $$\int_{a}^{b}$$f (x) dx

Case 3: The curve is part above and part below the x-axis

You have to split the problem up and consider the two parts separately. Use case 1 to deal with the section above the x-axis and case 2 to deal with the section below the x-axis.

Example 6..14   Find the area bounded by the curve y = sin x, the x-axis and the lines x = 0 and x = 2$\pi$.

Solution Draw a picture so that you can see what is happening.

From 0 to $\pi$ the curve is above the axis and from $\pi$ to 2$\pi$ it is below.

Above the axis we have

$$\int_{0}^{{\pi}}$$ Area 

$$\left[\vphantom{-\cos x}\right. \left.\vphantom{-\cos x}\right]_{0}^{{\pi}}$$

$$\pi$$ = 2

And below we have

$$\int_{{\pi}}^{{2\pi}}$$ Area 

$$\left[\vphantom{\cos x}\right. \left.\vphantom{\cos x}\right]_{{\pi}}^{{2\pi}}$$

$$\pi \pi$$ = 2

Therefore the total area is 4.

Note what would have happened had we not split the area into two sections. The two parts would have cancelled, giving us an area of zero -- which is clearly silly. In this case we could have noted, by symmetry, that the required area = 2$$\int^{{\pi}}_{0}$$sin x dx = 4.

The same ``assemble the bits'' approach works on areas between curves.

Example 6..15   Find the area between the curve y = sin x and the line y = $${\frac{{2x}}{{\pi}}}$$ and above the x-axis.

Solution Again the best way to begin is by drawing a picture.

Here A = $$\left(\vphantom{\frac{\pi}{2},1}\right.$$$${\frac{{\pi}}{{2}}}$$, 1$$\left.\vphantom{\frac{\pi}{2},1}\right)$$ and lies on both y = sin x and y = $${\frac{{2x}}{{\pi}}}$$.

The next step is to work out where the line and curve cross. This will give us the limits for the integrals.

In this case they cross when x = 0 and when x = 1.

From the picture we can see that the area we want is the area under the curve minus that under the line (shaded). Therefore

$$\int_{0}^{{\frac{\pi}{2}}} \int_{0}^{{\frac{\pi}{2}}} {\frac{{2x}}{{\pi}}}$$ Area 

$$\left[\vphantom{-\cos x}\right. \left.\vphantom{-\cos x}\right]_{0}^{{\frac{\pi}{2}}} \left[\vphantom{\frac{x^2}{\pi}}\right. {\frac{{x^2}}{{\pi}}} \left.\vphantom{\frac{x^2}{\pi}}\right]_{0}^{{\frac{\pi}{2}}}$$

$${\frac{{\pi^2}}{{4}}} {\frac{{1}}{{\pi}}}$$

$${\frac{{\pi}}{{4}}}$$

Example 6..16   Find the area between the curve y = x² and the line y = 2.

Example 6..17   Find the area between the curves y = x² and y = x⁴.