Chapter 4: Integration
Section 4.4: Integration by Substitution
Example 4.4.8
Prove that ∫−aafx ⅆx=2∫0afx ⅆx for any even function f.
Solution
Mathematical Solution
The assumption that f is even means f−x=fx.
Split the interval of integration at x=0, that is, at the center of symmetry.
∫−aafx ⅆx=∫−a0fx ⅆx+∫0afx ⅆx
In the first integral on the right, make the change of variables x=−t so that dx=−dt and x=−a⇒t=a.
∫−aafx ⅆx=−∫a0f−t ⅆt+∫0afx ⅆx
In the first integral on the right, reverse the order of integration, thereby changing the sign of the integral.
∫−aafx ⅆx=∫0af−t ⅆt+∫0afx ⅆx
In the first integral on the right, replace f−t with ft since f is an even function.
∫−aafx ⅆx=∫0aft ⅆt+∫0afx ⅆx
In the first integral on the right, t is a "dummy" variable, and can be replaced with any other letter. So, change t to x.
∫−aafx ⅆx=∫0afx ⅆx+∫0afx ⅆx
Arithmetic.
∫−aafx ⅆx=2∫0afx ⅆx
Maple Solution
Load the IntegrationTools package.
withIntegrationTools:
Control-drag the definite integral. Context Panel: Assign to a Name≻q
∫−aafx ⅆx→assign to a nameq
Apply to Split command to break the interval of integration at x=0.
Splitq,0
∫−a0f⁡xⅆx+∫0af⁡xⅆx
Control-drag the first summand . Context Panel: Assign to a Name≻L1
∫−a0fxⅆx→assign to a nameL1
Control-drag the second summand. Context Panel: Assign to a Name≻R
∫0afxⅆx→assign to a nameR
Apply the Change command to L1, changing the integration variable from x to −t.
Assign the result to the name L2.
L2≔ChangeL1,x=−t
−∫a0f⁡−tⅆt
Apply the Flip command to L2, thereby reversing the direction and limits of integration.
Assign the result to the name L3.
L3≔FlipL2
∫0af⁡−tⅆt
Apply the Change command to L3, changing the integration variable from t to x.
Assign the result to the name L4.
L4≔ChangeL3,t=x
∫0af⁡−xⅆx
Expression palette: Evaluation template Replace f−x with fx. (Recall that f is even, so f−x=fx.)
L4x=a|f(x)f−x=fx = ∫0af⁡xⅆx→assign to a nameL5
Add the modified first summand to the second.
Context Panel: Simplify≻Simplify
L5+R = 2⁢∫0af⁡xⅆx
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