# 4. The Definite Integral

by M. Bourne

In the last section, we used the expression

`int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a)`

to find the area under a curve.

`F(x)` is the integral of `f(x)`;`F(b)` is the value of the integral at the upper limit, `x = b`; and

`F(a)` is the value of the integral at the lower limit, `x = a`.

This expression is called a **definite integral.** Note that it does not involve a **constant of
integration** and it gives us a definite value (a number) at
the end of the calculation.

See more about the above expression in Fundamental Theorem of Calculus. It contains an applet where you can explore this concept.

We will use definite integrals to solve many practical problems. First, we see how to calculate definite integrals.

### Example 1

Evaluate `int_1^5(3x^2+4x+1)dx`

Note that our final answer is a **number** and does not involve "**+ K**". We are now dealing with **definite **integrals.

### Example 2

Evaluate `int_4^9(2x+3sqrtx)dx`

### Example 3

Evaluate `int_1.2^1.6(5+6/x^4)dx`

## Definite Integrals and Substitution

Recall the substitution formula for integration:

`int u^n du=(u^(n+1))/(n+1)+K` (if `n ≠ -1`)

When we substitute, we are changing the variable, so we cannot use the same upper and lower limits. We can either:

- Do the problem as an
**indefinite integral**first, then use upper and lower limits later - Do the problem throughout using the new variable and the new upper and lower limits
- Show the correct variable for the upper and lower limit during the substitution phase.

We will be using the third of these possibilities.

### Example 4

Find

`int_-1^0x^3(1-2x^4)^3dx`

using a substitution.

## Application: Work

Einstein riding his bicycle.

In
physics, **work** is done when a force acting upon an object
causes a displacement. (For example, riding a bicycle.)

If the force is not constant, we must use **integration**
to find the work done.

We use

`W=int_a^bF(x)dx`

where
*F*(*x*) is the variable force.

### Example 5

Find the work done if
a force `F(x)=sqrt(2x-1)` is acting on an object and moves it from *x* = 1 to *x* = 5.

For more on **work**, see: 7. Work by a Variable Force.

## Application: Average Value

The **average value** of a function
*f*(*x*) in the region *x = a* to *x = b* is given by:

`"Average"=(int_a^bf(x)dx)/(b-a)`

### Example 6

Find the average value of *x*(3*x*^{2} - 1)^{3} from 0 to
1.

Here is the graph of the situation:

## Application: Displacement

If we know the expression, *v*, for
velocity in terms of *t*, the time, we can find the **displacement** (written *s*) of a moving object
from time *t *=* a* to time *t *=* b * by integration, as follows:

`s=int_a^bv\ dt`

### Example 7

Find the displacement of
an object from *t* = 2 to *t* = 3, if the velocity of
the object at time *t* is given by

`v=(t^2+1)/((t^3+3t)^2`

See more on: displacement, velocity and acceleration as applications of integration.

**NOTE 1:** As you can see from the above applications of work, average value and displacement, the definite integral can be used to find more than just areas under curves.

**NOTE 2: **The definite integral **only** gives us an **area** when the whole of the curve is **above the x-axis** in the
region from *x = a* to *x = b.* If this is not the
case, we have to break it up into individual sections. See more at Area Under a Curve.

We now examine a definite integral that we cannot solve using substitution.

## Not every integral can be integrated using substitution...

Consider this question.

Find: `int_0^1sqrt(x^2+1)\ dx`

**Attempted solution:**

We try to put `u = x^2+ 1`.

Then we find the differential:

`du = 2x\ dx`

But the question **does not have** "`2x\ dx`" (it only has "`dx`") so
we cannot replace anything in the question with "`du`" properly. This means we can't solve it using any of the integration methods used
above. (**Note: **This question **can** be done using Trigonometric Substitution,
however, but we don't meet trigonometric substitution until later.)

**Explanation: **Let's think about if the question said this instead:

`int_0^1sqrt(x^2+1)(2x)dx`

This time, there **is** a "`2x\ dx`" involved in the question and so we would be able to let:

`u = x^2+ 1`

Then we would find the differential:

`du = 2x\ dx`

Then we could proceed to find the integral like we did in the examples above, by replacing `2x\ dx` with `du`* *and the square root part with `sqrt u`.

However, the problem `int_0^1sqrt(x^2+1)\ dx` does **not** have a "`2x`" outside of the square root so I cannot use the "`u`" substitution.

For now, we need to use **numerical approaches** to evaluate the integral

`int_0^1sqrt(x^2+1)\ dx`

(**Note: **Historically, all definite integrals were approximated using numerical methods before Newton and Leibniz developed the integration methods we have learned so far in this chapter.)

We can use two different numerical methods for evaluating an integral:

We meet these methods in the next 2 sections.

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