Calculus Of One Real Variable – By Pheng Kim Ving

10.2 
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In the previous section, Section
10.1, we discussed integration by inspection. There are lots of
integrals that are hard to evaluate
or can't be evaluated by inspection but that can each be found by an
integration technique or method. This section and several
subsequent ones present some integration techniques.
1. The Method Of Substitution 
Calculate this indefinite integral:
Solution
Let u = x^{2} – 1. Then du = 2x
dx. So:
EOS
We have 2x(x^{2} – 1)^{10} dx = (x^{2} – 1)^{10} (2x dx) = u^{10} du.
Recall that du
is the differential of u
(see Section
4.3 Definitions 2.1): du
= u'(x) dx = 2x
dx. In the integral, we spot
the factors 2x and x^{2} – 1, and we know that 2x is the derivative of x^{2} – 1, so the differential d(x^{2} – 1) of x^{2} – 1 has the
factor 2x in it: d(x^{2} – 1) = 2x
dx. Thus we substitute u = x^{2} – 1, calculate du,
and transform the integral in x
into one in
u. The integral in u is ready for an integration
formula to be applied to it. After finding the integral in u, we have to return
to the original variable, x
in this case, in our answer.
We substitute u
= x^{2} – 1. The
technique used is therefore called the method of substitution. If we
want to check to
see that our answer is correct, then we just differentiate it:
Compute:
Let u = x – 1, so that du = dx. Then:
EOS
Here d(x – 1) = 1 . dx = dx. So we substitute u = x – 1.
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2. The Method Of Substitution And The Chain Rule 
^{{2.1}} Example
1.1.
The method of substitution is valid because it's derived from the chain rule.
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3. Difference By A Constant Factor 
Evaluate:
Solution
1
Let u = x^{2} – 1. Then du = 2x
dx, so that x dx = (1/2) du.
Thus:
EOS
Solution 2
Let u = x^{2} – 1. Then du = 2x
dx. Thus:
EOS
EOS
In this example we have x instead of 2x,
so 1 term, x, isn't
exactly the derivative, 2x,
of the other, x^{2} – 1, but the
difference is only by a constant factor, namely 2. The difference by a constant
factor can be removed easily without
making the integral more complicated, as shown in each of the solutions.
The bypassing of explicit substitution as done in Solution 3 can be applied to any integral formula.
Find:
EOS
We think of x^{3} as u and write the integrand in
the form e^{u}
du so that we can apply the
integral formula for exponential
functions. In general this can easily be done when the substitution and the
formation of du in terms
of x and dx can
easily be performed mentally.
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4. Integrals Involving Radicals 
Calculate:
Solution
1
Let u = 5 – x^{2}. Then du = – 2x
dx, so that x dx = (–1/2) du.
Thus:
EOS
We can eliminate the radical, as shown in Solution 2 below.
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5. Handling The Definite Integrals 
Compute this definite integral:
EOS
We must write the limits of integration as x = 0 and x = 2 whenever the variable of integration is u, not just as 0 and 2,
because it's x, not u, that goes from 0 to 2. Here
we keep the xlimits of
integration, so we must return to x
before
substituting in the limits. Another approach is shown in Solution 2 below.
Solution
2
Let u = x^{3} + 1. Then du = 3x^{2} dx, so that x^{2} dx = (1/3) du.
When x = 0 we
have u = 0^{3} + 1 = 1, and
when x = 2 we
have u = 2^{3} + 1 = 9. Thus:
EOS
Here we transform the xlimits into the ulimits, and we evaluate the integral as soon as
we've found it in terms of u
using these ulimits.
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6. When The Method Of Substitution Is Likely To Work 
1. Calculate the following indefinite integrals.
Solution
a. Let u = 3x^{4} + 5. Then du = 12x^{3} dx, so x^{3} dx = (1/12) du. Thus:
2. Compute the following indefinite integrals.
Solution
a. Let u = s^{2}. Then du = 2s ds, so s ds = (1/2) du. Thus:
3. Show that:
Solution
We have 10 + 6x + x^{2} = 1 + (x + 3)^{2}. Let u = x + 3. So du = dx. Thus:
4. Evaluate the definite integral:
handling the
limits of integration in 2 ways:
a. Keep the xlimits.
b. Change to the limits for the substitution variable.
Solution
5. Find the area of the plane region bounded by
the graph of y = x/(x^{4} + 16), the xaxis,
the yaxis, and
the vertical line
x
= 2.
Solution
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