Finding the Area of the Graph

[ielle88]

Good day!

I am having a hard time answering these questions. Will you guys help me with this? I appreciate all of your replies. Thank you!

1. Find the area enclosed by the graph of f(x) = x^2 −4, the x−axis, and the line x = 0 and x = 4

2. Find the area in the first quadrant enclosed by the graph of y = f(x)= sinx, y=g(x)=cosx , and the y−axis, 0 ≤ x ≤ π

 

[Steven G]

This is a help forum! You need to inform us on what type of help you need. Did you graph the four functions? Can you post back with your graph and any work you have done so we know what type of help you need?

 

[nasi112]

For [MATH]1[/MATH]
This is the area that you want to find. Green Color.



When you integrate from [MATH]x = 0[/MATH] to [MATH]x = 2[/MATH], you will get a negative area. Use the absolute value to make it positive. Then, integrate from [MATH]x = 2[/MATH] to [MATH]x = 4[/MATH], you will get a positive area.

The required area is Area 1 + Area 2

 

[ielle88]

I'm sorry I am still new here.

This is the graph for the first question




I am confused about what area should I compute. Since the question only asks for the enclosed I thought that the Area I was looking for is the x = 0 to x=2. However, there is a given from x=0 to x=4. So I do not know if I need to get the total area?





For the second question I really do not understand it. Like what is the first quadrant it refers to? And how to solve with the reference of y
Here is the graph for question 2

 

[ielle88]

OMYHGAD THANK YOU SO MUCH!!! I was really confused about what area I was looking for. This clarifies it Thank youuu

 

[nasi112]

For [MATH]2[/MATH]
First quadrant means this,



The required area is this. The green color.



You will integrate from [MATH]x = 0[/MATH] to [MATH]x = \frac{\pi}{4}[/MATH]

 

[Steven G]

For the 1st problem you only drew one graph! As I said, there are four functions which you should draw. If you do, then you'll see the bounded area. The 4 functions are f(x) = x^2 −4, y=0 (the x−axis), x = 0 and x = 4. Just because some of the functions are not complicated doesn't mean that you do not draw it.

Here is a problem for you to thing about. Find the area bounded by y=5, y=2, x=4 and x=8. You can easily do this without calculus.

 

[nasi112]

For the 1st problem you only drew one graph! As I said, there are four functions which you should draw. If you do, then you'll see the bounded area. The 4 functions are f(x) = x^2 −4, y=0 (the x−axis), x = 0 and x = 4. Just because some of the functions are not complicated doesn't mean that you do not draw it.

Here is a problem for you to thing about. Find the area bounded by y=5, y=2, x=4 and x=8. You can easily do this without calculus.

Jomo. I have a question for you.

Can you prove that the area bounded by [MATH]y = 5, y = 2, x = 4, x = 8[/MATH] by using Calculus is equal to the area by not using Calculus?

 

[Deleted member 4993]

Jomo. I have a question for you.

Can you prove that the area bounded by [MATH]y = 5, y = 2, x = 4, x = 8[/MATH] by using Calculus is equal to the area by not using Calculus?

Refer to response #2 !!

 

[nasi112]

Refer to response #2 !!

It is a question for you Khan if you wanna answer it. Because I have a doubt that they are equal to each other.

 

[pka]

#1: $\displaystyle\int_0^2 {(4 - {x^2})dx} + \int_2^4 {({x^2} - 4)dx}=~?$

 

[pka]

#2: $\displaystyle 2\int_0^{\frac{\pi }{4}} {\left( {\cos (x) - \sin (x)} \right)dx}$ why?

 

[Deleted member 4993]

Jomo. I have a question for you.

Can you prove that the area bounded by [MATH]y = 5, y = 2, x = 4, x = 8[/MATH] by using Calculus is equal to the area by not using Calculus?

Without Calculus, you should use graphical method and get the area to be 12.

Now you show us the method by calculus and get a different number!!

 

[Steven G]

Jomo. I have a question for you.

Can you prove that the area bounded by [MATH]y = 5, y = 2, x = 4, x = 8[/MATH] by using Calculus is equal to the area by not using Calculus?

Yes, I can solve for the area using Calculus. Just graph the bounded area. If you do, then you'll see two ways to find the area using integrals.

 

[Steven G]

It is a question for you Khan if you wanna answer it. Because I have a doubt that they are equal to each other.

If the area is different depending on the method you use, then mathematics is flawed.

 

[nasi112]

#2: $\displaystyle 2\int_0^{\frac{\pi }{4}} {\left( {\cos (x) - \sin (x)} \right)dx}$ why?

Why are you doubling the required region?

Without Calculus, you should use graphical method and get the area to be 12.

Now you show us the method by calculus and get a different number!!

But it is a question for you, not me.

Yes, I can solve for the area using Calculus. Just graph the bounded area. If you do, then you'll see two ways to find the area using integrals.

I don't see any of your solutions.

 

[HallsofIvy]

It is a question for you Khan if you wanna answer it. Because I have a doubt that they are equal to each other.

That's a strange question! Why would you think the area of a region depends upon what method you use to find it?
Or, for that matter whether anyone finds that area at all!

 

[Steven G]

Yes, you got me. Using integrals you get one answer and without integrals you get another answer for the same area!

 

[lookagain]

Yes, you got me. Using integrals you get one answer and without integrals you get another answer for the same area!

You are referring to nasi112, correct?

The area bounded by y = 5, y = 2, x = 4, x = 8

"Top curve" minus the "bottom curve": y = 5 - 2 = 3

The integrand is 3. The antiderivative is 3x.

You're integrating from x = 4 to x = 8.

3(8) - 3(4) = 24 - 12 = 12, which is also the method done graphically.

 

[Steven G]

You are referring to nasi112, correct?

The area bounded by y = 5, y = 2, x = 4, x = 8

"Top curve" minus the "bottom curve": y = 5 - 2 = 3

The integrand is 3. The antiderivative is 3x.

You're integrating from x = 4 to x = 8.

3(8) - 3(4) = 24 - 12 = 12, which is also the method done graphically.

Of course I know the answer is 12 regardless of the method used.