Problem : Compute
sin2(x)cos(x)dx .
Making the change of variables u = sin(x) , the integral becomes
u 2 du =
+C
Substituting the definition of u , we obtain (sin3(x)/3 + C . It is easy to check that this has
derivative equal to sin2(x)cos(x) .
Problem :
Problem : Compute
x sin(x)dx .
Letting f (x) = x and g(x) = sin(x) in the formula for integration by parts, we have
=
x sin(x)dx
Problem : Find
- x cos(x) (- cos(x)dx
= - x cos(x) + sin(x) + C
x 2 e x dx .
Letting f (x) = x 2 and g(x) = e x , we have
=
x
We have not yet discussed how to integrate rational functions (recall that a rational function
is a function of the form f (x)/g(x) , where f , g are polynomials). The method that allows us to
do so, in certain cases, is called partial fraction decomposit
The natural rules for the definite integral of sums and constant multiplies of functions, i.e.
sumrule, constmult
(f (x) + g(x)dx =
f (x)dx +
cf (x)dx = c
g(x)dx
f (x)dx
follow (by the Fundamental Theorem of Calculus) from the similar rules for antideriva
Problem : Compute
2 cos(x) - 3e x
dx .
Use the addition and multiplication by a constant rules for integration:
=
2 cos(x) - 3e x
dx
2
cos(x)dx - 3
= 2 sin(x) - 3e x + C
e x dx
where C is an arbitrary constant.
The change of variables method for antidiffe
Recall that the area below the graph of the function f (x) from a to b is the definite integral
f (x)dx
where area counts as negative when f (x) < 0 . If the function f(x) takes on both positive and
negative values in the interval[a, b] , and we want to c
The application of integrals to the computation of areas in the plane can be extended to the
computation of certain volumes in space, namely those of solids of revolution. A solid of
revolution arises from revolving the region below the graph of a functio
Problem : Find the area of the region in the plane between the graphs of f (x) = | x|and g(x)
=-x2+6.
The intersection points of the graphs are at x = 2 , and - x 2 +6| x| on the interval[- 2, 2] , so
the area of the relevant region is equal to
=
(- x + 6
The integration by parts method comes from the product rule for derivatives. Given two
functions f , g , the product rule states that
[f (x)g(x)] = f'(x)g(x) + f (x)g'(x)
As usual, equating antiderivatives of these two expressions that agree at one point,
We have already seen that, in order to be able to compute definite integrals, it is enough to
be able to compute indefinite integrals (or antiderivatives). While for some functions, an
antiderivative can be guessed fairly easily (for example,
2 cos(2x)dx