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:
Let u = x2 – 1. Then du = 2x dx. So:
We have 2x(x2 – 1)10 dx = (x2 – 1)10 (2x dx) = u10 du.
Recall that du
is the differential of u
(see ): du
= u'(x) dx = 2x
dx. In the integral, we spot
the factors 2x and x2 – 1, and we know that 2x is the derivative of x2 – 1, so the differential d(x2 – 1) of x2 – 1 has the
factor 2x in it: d(x2 – 1) = 2x dx. Thus we substitute u = x2 – 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
= x2 – 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:
Let u = x – 1, so that du = dx. Then:
Here d(x – 1) = 1 . dx = dx. So we substitute u = x – 1.
2. The Method Of Substitution And The Chain Rule
The method of substitution is valid because it's derived from the chain rule.
3. Difference By A Constant Factor
Let u = x2 – 1. Then du = 2x dx, so that x dx = (1/2) du. Thus:
Let u = x2 – 1. Then du = 2x dx. Thus:
In this example we have x instead of 2x,
so 1 term, x, isn't
exactly the derivative, 2x,
of the other, x2 – 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.
We think of x3 as u and write the integrand in
the form eu
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.
4. Integrals Involving Radicals
Let u = 5 – x2. Then du = – 2x dx, so that x dx = (–1/2) du. Thus:
We can eliminate the radical, as shown in Solution 2 below.
5. Handling The Definite Integrals
Compute this definite integral:
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 x-limits of integration, so we must return to x before
substituting in the limits. Another approach is shown in Solution 2 below.
Let u = x3 + 1. Then du = 3x2 dx, so that x2 dx = (1/3) du. When x = 0 we have u = 03 + 1 = 1, and when x = 2 we
have u = 23 + 1 = 9. Thus:
Here we transform the x-limits into the u-limits, and we evaluate the integral as soon as
we've found it in terms of u
using these u-limits.
6. When The Method Of Substitution Is Likely To Work
1. Calculate the following indefinite integrals.
a. Let u = 3x4 + 5. Then du = 12x3 dx, so x3 dx = (1/12) du. Thus:
2. Compute the following indefinite integrals.
a. Let u = s2. Then du = 2s ds, so s ds = (1/2) du. Thus:
3. Show that:
We have 10 + 6x + x2 = 1 + (x + 3)2. Let u = x + 3. So du = dx. Thus:
4. Evaluate the definite integral:
limits of integration in 2 ways:
a. Keep the x-limits.
b. Change to the limits for the substitution variable.
5. Find the area of the plane region bounded by
the graph of y = x/(x4 + 16), the x-axis,
the y-axis, and
the vertical line
x = 2.