**Definite & Indefinite Integrals
**

There are 2 types of integral - (i) *Indefinite*, in which
we *aren't* given the limits of integration, *i.e. x*=*a*
to *x*=*b*, and (ii) *Definite*, in which we are
told *a* and *b* and so we can calculate a value for
an area.

**Indefinite integrals
**

If the differential of *x*^{3} is 3*x*^{2},
then

3*x*^{2}.*dx* = *x*^{3}

But 3*x*^{2} is the differential of *x*^{3}-1
and *x*^{3}+8, *etc*. so that this reversal
is not unique - we've 'lost' the *constant*!

So in general, 3*x*^{2} is the differential
of *x*^{3}+*k* where *k* is any
constant - known as the *'constant of integration'*.

We write this as: 3*x*^{2}.*dx*
= *x*^{3} + *k
*

**Integration 'Magic' Formula**

Since integration is the reverse of differentiation, for any
polynomial *y*(*x*) = *x ^{n}*, the integral is given
by:

In words: Add one to the power, then divide by the new power. Then add *k*.

*Examples *

1. *x*^{2}.*dx* =
+ *k
*

2. 20*x*^{4}.*dx*
= 4*x*^{5} + *k
*

3. 7x.*dx* =
+ *k
*

4. *a*.*dx* = *ax* + *k*

**Roots follow the same rule**

5. .*dx* = *x*^{½}.*dx*
= + *k
*

**Inverse powers follow the same rule
**

6. .*dx* =
*x*^{-3}.*dx* =
= - + *k
*

This is true so long as the exponent is *not* -1.
cannot be calculated using this
formula because we get a divide by zero error.

**We can use other variables as well
**

7. 2*m*^{2}.*dm*
= + *k
*

8. 5 .dλ =
+ *k
*

9. .dθ = -½.dθ = +*k
*

**Sums of terms
**

Just as in differentiation, a function can by integrated term by term, and we only need one constant of integration.

10. 3*x*^{2} + 7*x.dx*
= 3*x*^{2} + 7*x* = *x*^{3} +
+ *k
*

11. +
+ + 4*x*^{3}.*dx*
= + *x*^{4} + *k*

**Definite Integrals
**

We now know how to integrate simple polynomials, but if we want
to use this technique to calculate *areas*, we need to know
the *limits* of integration. If we specify the limits *x
= a* to *x = b*, we call the integral a *definite integral.
*

To solve a definite integral, we first integrate the function
as before, then feed in the 2 values of the limits. Subtracting
one from the other gives the *area*.

*Example
*

1. What is the area under the curve *y*(*x*) = 2*x*^{2}
between *x*=1 and *x*=3? (Note: this is the same problem
we did graphically earlier).

Area = we write the limits at the top and bottom of the integration sign

= we use square brackets to
indicate we've calculated the indefinite integral

= (18 + *k*) - (2/3 + *k*)
feed in the larger value, then the smaller, and subtract
the two.

= 18 - 2/3

= __17.33 sq. units__ (compare the approximate value we got graphically of 17¾).

**Note: **the *k*'s cancel. So when we evaluate a *definite
integral* we can ignore the constant of integration.

2. What is the area under the curve *y*(*x*) = 2*x*^{3}
- 6*x* between *x* = -1 and *x* = 0?

A =

=

= (0) - (½ - 3) = __2½ sq.units__

3. What is the enthalpy of a gas at 20 K given that its heat capacity
as a function of temperature is given by *C* = 2*T*^{2},
over the range *T* = 0 K to 30 K?

The enthalpy of a gas, *H*, is given by the area under the curve
of heat capacity vs temperature. In most cases, we approximate
it by saying that the heat capacity doesn't change much with *T*,
so is in fact a constant. In this case the enthalpy is just given
by *H* = *C*(*T*_{2}-*T*_{1}) since

= [*C.T*] = *C*(*T*_{2}
- *T*_{1}).

However in this question, we are asked for a more accurate answer,
and are told *C* is not constant, it's a function of *T*.

So *H* =

= = = (16000 / 3) - 0

= 5333.3

= __5.3 kJ mol__^{-1}

4. What is the area under the curve *y*(*x*) =
between *x* = 1 and *x* = infinity?

A =

=

= [-2*x*^{-1}] =

= (0) - (-2) = __2 sq.units__.

Consider the function *y*(*x*) = 2*x* within the
limits *x* = -2 to +1.

A = = [*x*^{2}]
= (1^{2}) - (-2^{2}) = **-3
units**

What does *negative area *mean?

The area *A*_{1} = ½ x 4 x 2 is below the *x*
axis and is counted as *-ve.*

The area *A*_{2} = ½ x 1 x 2 is above
the *x* axis and is counted at *+ve.
*

Therefore it is always a good idea to sketch a curve before you integrate, to see if it goes -ve anywhere between the limits.

For any function for which the differential has been established, reversal of the process gives the integral.

*Examples *learn these!

1. **Exponential Functions**

e* ^{x}.dx* = e

e* ^{ax}*.

2. **Logarithmic Functions**

.*dx* = ln *x* + *k* (this is the one we cannot do using the 'magic formula').

ln *x.dx* = *x*.(ln *x* - 1) + *k*

3. **Trigonometrical Functions**

cos *x.dx* = sin *x* + *k*

sin *x.dx* = - cos *x* + *k*

tan *x.dx* = - ln (cos *x*) + k

*Examples*

1. What is the area under the curve *y*(*x*) = 3e^{-5x}
from *x* = 1 to *x *= infinity ?

*A* =

=

= (0) - (-0.004) = __0.004 sq.units__

2. What is the area under the curve *y*(θ) = 3sin θ between θ = 0 and ?

*A* = 3 sin θ.*d*θ

= [-3 cos ]

= (-3 x 0.876) - (-3)

**= 0.373 sq.units**

3. Integrate *y*(*x*) = 4cos *x* + 3e^{2x} -

4cos *x* + 3exp(2*x*) - .*dx*

= 4sin *x* + e^{2x}- 5ln *x* + *k*

End of Course.