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I suppose you could solve this algebraically, but apparently, there is a way to solve this without doing anything to the red area to do this question. Can someone explain how?

Jim
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  • I suppose we would need values(of the blue area). – Shubham May 24 '14 at 10:31
  • This is what the question says. No values given. – Jim May 24 '14 at 10:32
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    The equation to be solved is $$ 3\color{blue}{\int_a^b x^n dx} = \color{red}{\int_{a^n}^{b^n}x^{1/n} dx} $$ but I hope (and believe) that there is some more elegant argument out there somewhere. – Arthur May 24 '14 at 10:37

5 Answers5

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The area covered by the red and blue area is $b\cdot b^n - a\cdot a^n$
As the red area is three times bigger than the blue one, the blue one is a quarter of the blue and red area, hence :

$$\int_a^b x^n = \frac{b^{n+1}-a^{n+1}}{4}$$ so $$\frac{b^{n+1}-a^{n+1}}{(n+1)}= \frac{b^{n+1}-a^{n+1}}{4}$$ so, if $b^{n+1}-a^{n+1}\neq 0$ $$(n+1)= 4$$ so $$n = 3$$

Fabien
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Hint: If you label the endpoints of the red region on the $y$ axis, they are at $a^n$ and $b^n$. That makes the area of little white rectangle near the origin $a\cdot a^n=a^{n+1}$. Thus you can write three equations in five unknowns:

$$R=3B$$ $$R+B=b^{n+1}-a^{n+1}$$ $$B=\int_a^b x^n\,dx$$

Barry Cipra
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B=blue area, R=red area.

$R+B=b\cdot b^{n}-a\cdot a^{n}=\\=b^{n+1}-a^{n+1}$

$R+B=3B+B=4B$

$4B=b^{n+1}-a^{n+1}$

$B=\int_{a}^{b}x^{n}dx=\frac{b^{n+1}}{n+1}-\frac{a^{n+1}}{n+1}$

poolpt
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The red, blue and white (the small rectangle) areas combined equal $b\cdot b^n=b^{n+1}$, which is the area of the big rectangle.

The white area is $a\cdot a^n=a^{n+1}$.

To calculate the blue area, you need to integrate between $a$ and $b$ resulting in $$\int_a^bx^ndx=\left[\frac{x^{n+1}}{n+1}\right]_a^b=\frac{b^{n+1}-a^{n+1}}{n+1}$$

As the red area is three times the blue area we have $$\text{Red area}=b^{n+1}-a^{n+1}-\frac{b^{n+1}-a^{n+1}}{n+1}=3\left(\frac{b^{n+1}-a^{n+1}}{n+1}\right)\\\Rightarrow b^{n+1}-a^{n+1}=4\left(\frac{b^{n+1}-a^{n+1}}{n+1}\right)\\\Rightarrow n = 3$$

Alijah Ahmed
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You may try this:

Let $A_B$ the area of the blue region and $A_R$ the area of the red one. Then, you can show that:

$$A_B = \int_a^b x^n \, \mathrm{d} x = \frac{b^{n+1} - a^{n+1}}{n+1}, \quad A_R = b b^n - A_B - a a^n,$$ so the condition $A_R = 3 A_B$ yields:

$$b\, b^n - A_B - a \, a^n = 3 A_B \Rightarrow b b^n - a a^n = \frac{4}{n+1} (b^{n+1} - a^{n+1}).$$ Note now that the LHS of the equation is just the same as the term between brackets, so you can solve for $n$.

Hope this helps.

Cheers!

Dmoreno
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