Draw a diagram to interpret this equation geometrially as an equality

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In summary, we discussed three problems involving integrals and functions. For the first problem, we proved the equality of areas for a continuous function on the real numbers. For the second problem, we used the substitution rule to show that the equality of areas holds for a translated function. Finally, for the third problem, we applied the substitution rule again to show that the equality of areas still holds for positive numbers a and b. Additionally, we discussed how to interpret these equations geometrically using diagrams.
  • #1
DivGradCurl
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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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  • #2
This looks fine to me.
Where do you need help?
 
  • #3
arildno said:
This looks fine to me.
Where do you need help?

Oh, good! Thanks for checking it out.

Other than that, I need to take care of the diagrams (problems 1 & 2). I'm not so sure how to handle those. It seems that the areas in #1 are symmetric with respect to the y-axis.
 
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  • #4
You're right about diagram 1.
For diagram 2, remember that f(x+c) represents a translation.
 
  • #5
arildno said:
You're right about diagram 1.
For diagram 2, remember that f(x+c) represents a translation.

I've got it. Thanks!
 

1. What does it mean to "draw a diagram to interpret an equation geometrially"?

Interpreting an equation geometrically means representing the equation in a visual form, such as a graph or diagram. This helps to better understand the relationship between the variables in the equation.

2. How can a diagram help in understanding an equation?

A diagram can provide a visual representation of the equation, making it easier to visualize the relationship between the variables and understand how they interact with each other.

3. What does it mean to interpret an equation as an equality?

An equation is a statement that shows the equality between two expressions. Interpreting an equation as an equality means understanding that the two sides of the equation are equal to each other.

4. What type of equations can be interpreted geometrically?

Most equations can be interpreted geometrically, as long as they have two or more variables that can be represented on a graph or diagram.

5. How can drawing a diagram help in solving an equation?

Drawing a diagram can help in solving an equation as it provides a visual representation of the equation, making it easier to understand and manipulate. It can also help in identifying patterns and relationships between the variables, which can aid in finding a solution.

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