INTEGRATION 1
1. Standard Results
The left-hand column lists derivatives encountered in units C1 and C3. The right-hand column rewrites these as integrals, which we will make use of in unit C4.
/ / / ** : this one isn’t strictly true, but more on this later!
Example 1 : Integrate the following.
a) b) c)
a) / / b) /c) /
Example 2 : Integrate the following.
a) b)
a) / / b) /C4 p83 Ex6A
2. Integration by Inspection
This is basically the chain rule in reverse. In unit C3 we used the chain rule to show that
The table below gives some examples.
y /So if we see an expression where the derivative of the function inside the brackets appears outside the brackets, we can integrate it straight away (by inspection).
Example 1 : Find .
Since 2 is the derivative of , we can integrate by inspection.
A quick check shows that if we differentiate we get our original function. Always check integrations done by inspection by re-differentiating!
Example 2 : Find .
The derivative of is , so to obtain this we rewrite the integral as follows.
Example 3 : Find .
The derivative of is , so to obtain this we take a factor of 2 outside the integral.
Example 4 : Find .
The derivative of is , so we take a factor of 3 outside the integral.
Example 5 : Find a) b)
a) / / b) /C4 p86 Ex 6B Q3bcfghi, 4d p94 Ex6E Q1cdf, 2afgh
3. Integration with Exponential Functions
In unit C3 we learned that remains the same when differentiated. Therefore, it will also remain the same when integrated, and so we have
Example 1 :Find, in terms of e, the area under the curve from to .
In unit C3, we used the chain rule to find a general rule for differentiating expressions of the form .
The table below gives some examples.
Reversing this result, we have integration by inspection.
Example 2 : Find the following integrals.
a) b) c)
We ‘set up’ the integrals by taking out a factor that leaves an expression of the form inside the integral.
a) / / b) / / c) /C4 p86 Ex 6B Q1bc, 2b, 4b p94 Ex 6E Q1g 2d
4. Integration with Logarithmic Functions
In unit C3, we learned that
This means that
In actual fact this is not strictly true, for reasons we will come to shortly. However, it will do for now!
Example1:Find the area under the curve from to .
We now look at a problem with our present result
Let use suppose we are integrating between, say –3 and –2. This integral represents the area bounded by the curve , the x-axis, and the lines and , labelled A in the diagram below.
...and now we have a problem because we cannot take the logarithm of a negative number.
However notice, by the symmetry of the graph, that the area A is the same as area B, but negative.
Now this is what we get if we just ignore the minus signs in the limits of the original integral. Therefore we have the true result
This also means that is the derivative (gradient function) of .
Some time spent considering the graphs below shows this makes sense.
Activity 4 : Plot the graph of on Autograph. Then use slow plot and gradient function, and check that the gradient function is the positive part of the graph. Then plot the graph of . Again use slow plot and gradient function, and check that the gradient function is the full graph.
Example2:Find the area between the curve , the x-axis and the lines and .
Example3:Find the equation of the curve which passes through (2, 3) and has gradient function
Integrating,
Substituting (2, 3),
And so the equation of the curve is
In unit C3, we used the chain rule to find a general rule for differentiating expressions of the form .
Reversing this result, we have integration by inspection.
Notice the modulus signs, for the same reasons as we discussed earlier.
Example 4 : Find the following integrals.
a) b) c)
Once again, we set up the integrals first.
a) / / b) /c) /
There is no need for modulus signs in a) as is always positive.
Example 5 : Find the integral of .
This expression is already in the form ‘f dash over f’, and so
Example 6 : Find the integral of .
This expression is already in the form ‘f dash over f’, and so
C4 p86 Ex 6B Q3adej p94 Ex 6E Q1abe, 2e
5. Integration with Trigonometric Functions
In unit C3, we used the chain rule to derive these results…
…and we can reverse these in order to integrate by inspection. We get integrals of the following form…
Example 1 : Find the following integrals.
a) b) c)
As with polynomial, exponential and logarithmic integrals met earlier in this topic, we need to do some ‘setting up’ of each integral first, to get it in the required form. Here, we need to ensure appears in front of the trigonometric function.
a) / / b) /c) /
Example 2 : Find the following integrals.
a) b) c)
We do these by considering what power of which trigonometric function we need, and then selecting the appropriate constant in front. Notice the ‘trick’ in part c).
a) / / b) /c) /
Example 3 : Find the following integrals.
a) b) c)
In a similar way, get the right power of the right trigonometric function, and worry about the constant afterwards.
a) / / b) /c) /
Example 4 : Find .
We recognise that we can manipulate this expression into the form .
Example5: Find the area underneath the graph of between and .
Finally, you will notice that we have not yet integrated tan, cot, sec or cosec. These integrals involve the use of , and sometimes in a very sneaky way…!
C4 p86 Ex 6B Q1 (not b), 2acde, 4ac p94 Ex 6E Q1hij, 2bcij
6. Using Trigonometric Identities in Integration
Standard trigonometric identities such as Pythagorean relationships, double-angle formulas and factor formulas (from unit C3) can help us do other trigonometric integrations. The first two examples illustrate the use of the identity
Example 1 : Find .
Note : a lot of students try this sort of thing…
…and then get stuck. Hopefully you should realise that using the Pythagorean identity makes us no better off!
Example 2 : Find .
For odd powers of sine and cosine, the Pythagorean identity is useful.
Example 3 : Find .
Other Pythagorean identities also come in handy.
Example 4 : Find .
Activity 2 : Three students integrated in these three ways.
Explain how three different answers can all be correct. You may find it helpful to use Autograph to draw the graphs of all three answers on the same axes.
Example 5 : Integrate the following…
a) b) c)
a) / Using a double-angle formulab) /
c) / Using the factor formula with ...
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7. Integration Using Partial Fractions
In unit C4, we have already used partial fractions to help us differentiate and find binomial expansions. We can also use them to integrate certain expressions.
Example 1 : Express in partial fractions. Hence integrate this function.
Comparing numerators,
Therefore we have
Example 2 : Express in partial fractions. Hence integrate this function.
We begin by factorising the denominator as far as possible.
We have three linear factors. So,
Comparing numerators (once the RHS has been written as a single fraction),
Therefore
Example 3 : Express in partial fractions, and hence integrate it.
Comparing numerators,
Equating coefficients of ,
Therefore
A useful trick with log integrals is to decompose improper algebraic fractions.
Example 4 : Integrate the following…
a) b) c)
a) / Decomposing the numerator,Note that you can also use algebraic long division.
1
x / − / 4 / x / + / 0
x / − / 4
4
And so…
/ c) /
Comparing numerators,
/ /
/ /
b) /
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8. Integration by Substitution (Change of Variable)
Consider the integral .
Clearly we can do this by inspection, by remembering that
In this case, and . Therefore
However, we can also use substitution. Here, we introduce a second variable u, and rewrite the integral in terms of u. The dx must also be eliminated, and this can be done by finding (although sometimes is more convenient).
So we have
which gives us the same answer as before.
Example 1 : Find .
This integration cannot be done by inspection, since the derivative of the contents of the brackets is 1, and we have an x outside. We could expand the brackets and integrate each term separately, but this would be tedious. Using substitution here saves time and effort.
The main use of substitution is to perform integrations that cannot be done any other way...
Example 2 : Find .
If the integral has limits, then we can change these too. This saves us the bother of rewriting the answer back in terms of x at the end and using the original limits.
Example 3 : Evaluate the integral
Using the substitution ,
Example 4 : Repeat the previous example with the substitution .
In this example we can differentiate implicitly to find a relationship between and .
...and we have the same answer as in example 3.
Example 5 : Evaluate
Example 6 : Use the substitution to evaluate the integral .
Very convenient!
Example 7 : Use the substitution to evaluate the integral
Example 8 : Use the substitution to evaluate the integral
Example9: Evaluate the integral below, leaving your answer in terms of e.
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9. Integration by Parts
The product rule for differentiation is
Integrating both sides with respect to x
Rearranging, we have
This is the formula for integration by parts, and is useful for integrating products of functions.
Example 1 : Find .
Activity1:This was actually example 1 from the previous section (integration by substitution). Check the answers are the same.
Example 2 : Find .
We cannot choose , since this would leave us with , and we would then need to integrate to get v - which we don’t know how to do (yet).
Example 3 : Find .
We can write this as the product of two functions thus :
The next example illustrates the use of limits with integration by parts (definite integrals).
Example 4 :Find .
Example 5 : Find .
We must now use integration by parts to evaluate .
Substituting, we have
Example6: Find .
Example 7 (old chestnut) : Find .
We must now use integration by parts to evaluate . We must use once more, otherwise we go back to where we started! (Try it and see…)
Substituting,
This may appear to be a dead end, but…
Example 8 : Evaluate, using integration by parts, .
Note that we could also have used substitution (which would probably have been easier!)
C4 p102 Ex 6G Topic Review : Integration 1