(adsbygoogle = window.adsbygoogle || []).push({}); I need some help with the following problems. Any help is highly appreciated.

1. If [tex]f[/tex] is continuous on [tex]\mathbb{R}[/tex], prove that

[tex]\int _a ^b f(-x) \: dx = \int _{-b} ^{-a} f(x) \: dx[/tex]

For the case where [tex]f(x) \geq 0[/tex] and [tex]0 < a < b[/tex], draw a diagram to interpret this equation geometrially as an equality of areas.

2. If [tex]f[/tex] is continuous on [tex]\mathbb{R}[/tex], prove that

[tex]\int _a ^b f(x + c) \: dx = \int _{a+c} ^{b+c} f(x) \: dx[/tex]

For the case where [tex]f(x) \geq 0[/tex], draw a diagram to interpret this equation geometrially as an equality of areas.

3. If [tex]a[/tex] and [tex]b[/tex] are positive numbers, show that

[tex]\int _0 ^1 x^a (1 - x) ^b \: dx = \int _0 ^1 x^b (1 - x) ^a \: dx[/tex]

Here is what I've got so far:

1. Consider the left-hand side

[tex]\int _a ^b f(-x) \: dx[/tex]

and apply the substitution rule:

[tex]u=-x \Rightarrow \frac{du}{dx} = -1 \Rightarrow dx = - du[/tex]

[tex]u(b)=-b[/tex]

[tex]u(a)=-a[/tex]

[tex]\int _a ^b f(-x) \: dx = -\int _{-a} ^{-b} f(u) \: du = \int _{-b} ^{-a} f(u) \: du = \int _{-b} ^{-a} f(x) \: dx[/tex]

2. Consider the left-hand side

[tex]\int _a ^b f(x + c) \: dx[/tex]

and apply the substitution rule:

[tex]u=x+c \Rightarrow \frac{du}{dx} = 1 \Rightarrow dx = du[/tex]

[tex]u(b)=b+c[/tex]

[tex]u(a)=a+c[/tex]

[tex]\int _a ^b f(x + c) \: dx = \int _{a+c} ^{b+c} f(u) \: du = \int _{a+c} ^{b+c} f(x) \: dx[/tex]

3. Consider the left-hand side

[tex]\int _0 ^1 x^a (1 - x) ^b \: dx[/tex]

and apply the substitution rule:

[tex]u=1-x \Rightarrow \frac{du}{dx} = -1 \Rightarrow dx = -du[/tex]

[tex]u(1)=0[/tex]

[tex]u(0)=1[/tex]

[tex]\int _0 ^1 x^a (1 - x) ^b \: dx = \int _1 ^0 u^b (1 - u) ^a \: du = \int _0 ^1 x^b (1 - x) ^a \: dx[/tex]

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# Homework Help: Draw a diagram to interpret this equation geometrially as an equality

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