Find the limit a such that the definite integral equals 8

[naaa00]

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.

 

[emailanmol]

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)

 

[naaa00]

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...

:(

 

[emailanmol]

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)

 

[emailanmol]

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.

 

[SammyS]

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.

 

 

[naaa00]

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!

 

[SammyS]

...

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."