Find the limit a such that the definite integral equals 8

In summary, the person wrote a summary of the conversation. They made a mistake when they stated that f(x)(i.e 2x-x^2) is positive for x less than 2. They plugged in x = (-1) and got the correct result. They then stated that for 0 < x < 2, f(x) = 2x-x^2. For all other values of x, f(x) = x2-2x. They also stated that f(x) is never negative.
  • #1
naaa00
91
0
Find the limit "a" such that the definite integral equals 8

Homework Statement



Hello!

the problem ask me to find for which a in R, does this holds:

2
∫ abs(2x - x^2) = 8
a

The Attempt at a Solution



Well, I see that:

f(x) { if x > 2, then f(x) is negative; if x < 2, then f(x) is positive.

I rewrited the integral and got: (the limits are those inside the brackets)

2
∫ (-2x + x^2) + [2,a]∫ (2x - x^2) = [-a,2] - ∫ (2x - x^2) + [2,a]∫ (2x - x^2) or
-a a
- ∫ (2x - x^2)
-a

I know that in such cases, odd functions cancel. I'm left with the even function. I continued and got:

(2/3) * a^3

and since the definite integral should equal to 8, I just solved for "a" and got: (2/3) * a^3 = 8 or

a = cuberoot(12).

But when I wanted to check if (2/3) * (12)^1/3

I didn't got the expected 8.

...

So it must be wrong.

I don't see the mistake in my answer.

Any suggestions?

Thanks.
 
Physics news on Phys.org
  • #2


You made a mistake where you stated f(x)(i.e 2x-x^2) is positive for x less than 2.

Plug in x = (-1)
 
  • #3


hello!

You are right. Yeah, I should have said from 0 <= x <= 2.

0
∫ (-2x + x^2) + [0,2]∫ (2x - x^2) = [-a,0] - ∫ (2x - x^2) + [0,2]∫ (2x - x^2)
-a

And then I ended with a cubic polynomial:

a^3 + 3a^2 -20 = 0

Which has negative discriminant...

and its real solution , which is

a = 2

Doesn't work...

:(
 
Last edited:
  • #4


Your expression is nearly right.You have made a small error.

(Hint:in the expression coeffient of one term(either a^3 or a^2 term should come up negative)
 
  • #5


Solving the sum is pretty easy.

Since the graph of abs|f(x)| will lie always above x axis, amd the integral fetches a positive value a<2 (cause if a>2 integral of any positive function for a to 2 will be negative)

If 2>a>0 you can remove || as it has no meaning.
If a<0, the you can break the integral from a to 0 integral of [-f(x)] {why did i multioly with minus?} and from 0 to 2[integral of f(x)] {why no minus here?} .

Solve the sum to be equal to 8.
 
  • #6


attachment.php?attachmentid=45178&stc=1&d=1331939495.gif


The above is the graph of f(x) = |2x - x2| .

For 0 < x < 2 , f(x) = 2x - x2 .

For all other values of x, f(x) = x2 - 2x .

By the way, f(x) is never negative.

 

Attachments

  • WAlph_absVal.gif
    WAlph_absVal.gif
    3.1 KB · Views: 678
  • #7


naaa00 said:

Homework Statement




(2/3) * a^3 = 8 or

a = cuberoot(12).

But when I wanted to check if (2/3) * (12)^1/3

I didn't got the expected 8.

Apparently I commited a mistake at the beginning (and all of the time) when I wanted to check the value of a. You see in the quote when doing the substituion

(2/3) * (12)^1/3

It is supposed to be (2/3) * (12), which is equal to 8. For some reason I was not canceling the cube root of 12.

But still I my first answer was sloppy. Actually the suggestions (specially the graph) helped me a lot! I tried solving it again and got the answer correct!

Thank you!
 
  • #8


naaa00 said:
...

Actually the suggestions (specially the graph) helped me a lot! I tried solving it again and got the answer correct!

Thank you!
You're welcome.

As they say, "A picture is worth a thousand words."
 

FAQ: Find the limit a such that the definite integral equals 8

What is the definition of a limit?

A limit is the value that a function approaches as the input value gets closer and closer to a particular value.

How do you find the limit of a function?

To find the limit of a function, you must evaluate the function at values closer and closer to the desired input value and see what value the function approaches.

What is the definite integral of a function?

The definite integral of a function is the area under the curve of the function, bounded by the specified limits of integration.

How do you evaluate a definite integral?

To evaluate a definite integral, you must first find the antiderivative of the function and then plug in the specified limits of integration and subtract the values to find the area under the curve.

How do you determine the value of a to make the definite integral equal to 8?

To determine the value of a, you must first set up the definite integral with the function and specified limits of integration. Then, you can use algebraic manipulation to solve for a and make the definite integral equal to 8.

Back
Top