Chapter 4: Integration
Section 4.4: Integration by Substitution
Prove that ∫−aafx ⅆx=2∫0afx ⅆx for any even function f.
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
∫−aafx ⅆx=2∫0afx ⅆx
Load the IntegrationTools package.
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.
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.
Apply the Flip command to L2, thereby reversing the direction and limits of integration.
Assign the result to the name L3.
Apply the Change command to L3, changing the integration variable from t to x.
Assign the result to the name L4.
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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