The Student Room Group

Integration Problem

The diagram shows the graph of, where a ] 1, [.
The area of the pink region is equal to the area of the blue region. Give two equations for a in terms of b, and hence give a in exact form and determine the size of the blue area.

Must need to integrate with respect to x and y.
So one is the integral of
Unparseable latex formula:

$x^2$


And one is the integral of
Unparseable latex formula:

$\sqrt x$



Equate the two and you get the answer - but I don't so I must be wrong, but how am I wrong?
It's not the integral of Sqrt(x), but the integral of Sqrt(y).

Then you integrate one with respect to y and the other as you would normally with respect to x.
Reply 2
Sure, from 1 to b, but I can't seem to avoid a = 1!! :frown:
Reply 3
Integrate both sides and put in a and b leaving:

1a{x2}dx=[x33]1a\displaystyle \int_1^a \{x^2\} dx = \left[ \frac{x^3}{3} \right]_1^a

b1{y}dy=[2y323]b1\displaystyle \int_b^1 \{\sqrt{y}\} dy = \left[ \frac{{2y^\frac{3}{2}}}{3}\right]_b^1

But it's wrong!!
Reply 4
Original post by Gmart
Integrate both sides and put in a and b leaving:

1a{x2}dx=[x33]1a\displaystyle \int_1^a \{x^2\} dx = \left[ \frac{x^3}{3} \right]_1^a

b1{y}dy=[2y323]b1\displaystyle \int_b^1 \{\sqrt{y}\} dy = \left[ \frac{{2y^\frac{3}{2}}}{3}\right]_b^1

But it's wrong!!


Why do you think it's wrong?

It might be helpful to note that b=a2b = a^2.
Reply 5
Because:
Unparseable latex formula:

$a = 1+\sqrt{3}


and the above just goes to a = 1
(edited 7 years ago)
Reply 6
Original post by Gmart
Because:
Unparseable latex formula:

$a = 1+\sqrt{3}


and the above just goes to a = 1


I think that the question is just wrong and no such region exists.
Original post by Gmart
Integrate both sides and put in a and b leaving:

1a{x2}dx=[x33]1a\displaystyle \int_1^a \{x^2\} dx = \left[ \frac{x^3}{3} \right]_1^a

b1{y}dy=[2y323]b1\displaystyle \int_b^1 \{\sqrt{y}\} dy = \left[ \frac{{2y^\frac{3}{2}}}{3}\right]_b^1

But it's wrong!!


Surely you should be integrating the second integral from 1 to b, not b to 1?
Reply 8
Original post by TimGB
Surely you should be integrating the second integral from 1 to b, not b to 1?


Yeah, I agree - but that still gets you a=1a=1.
Original post by Zacken
Yeah, I agree - but that still gets you a=1a=1.


Hmm. That's unfortunate. However, with a bit of geometry you can prove that no such a exists greater than 1.
Original post by Zacken
I think that the question is just wrong and no such region exists.


This diagram should help explain why the question is impossible.
Reply 11
Original post by TimGB
This diagram should help explain why the question is impossible.


Oh, that's neat - I wouldn't have thought about that secant line from A to B!

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