Chapter 4: Integration
Section 4.4: Integration by Substitution
Table 4.4.1 lists the rule for making a substitution, or change of variable, in an integral.
Substitution Rule for Integrals
f and g satisfy suitable conditions
∫fx ⅆx= ∫fgyg′y ⅆy=Fy=Fg−1x
∫abfx ⅆx=∫g−1ag−1bfgy ⅆy
Table 4.4.1 Substitution in definite and indefinite integrals
If Fy is the antiderivative obtained when the substitution x=gy is made in the indefinite integral, then Fg−1x is an indefinite integral in terms of the original variable, x; correspondingly, the value of the definite integral is given by Fg−1b−Fg−1a = Fg−1x|ab.
Alternate forms for the substitution are y=hx and ry=sx. In either of these cases, obtain dx by solving explicitly for x and differentiating, or by differentiating implicitly. No matter how the substitution rule is given, an explicit solution for y=yx is needed for expressing the indefinite integral in terms of x.
Table 4.4.2 details the value of a definite integral of an odd or even function taken over a symmetric interval such as −a,a.
odd: f−x= −fx
Table 4.4.2 Definite integral of an odd or even function over a symmetric interval
In each case, the region shaded in yellow to the left of the y-axis represents the value of the definite integral on −a,0; the region shaded in red to the right of the y-axis, the value of the definite integral on 0,a. For the even function, the yellow and red regions match exactly, but for the odd function, the yellow and red regions match in magnitude, but differ in sign.
Evaluate the indefinite integral ∫3 x−2ⅆx.
Evaluate the indefinite integral ∫2 x+18 ⅆx.
Evaluate the indefinite integral ∫x 1−3 x2 ⅆx.
Evaluate the indefinite integral ∫x3 x2+3 ⅆx.
Evaluate the definite integral ∫23x3x2+3 ⅆx.
Verify that ∫−ππsinx ⅆx=0, and that ∫−π/2π/2cosx ⅆx=2∫0π/2cosx ⅆx.
Prove that ∫−aafx ⅆx=0 for any odd function f.
Prove that ∫−aafx ⅆx=2∫0afx ⅆx for any even function f.
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