Area of graph $y=x^n$ 
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?
 A: 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
$$n = 3$$
A: 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:
$$R=3B$$
$$R+B=b^{n+1}-a^{n+1}$$
$$B=\int_a^b x^n\,dx$$
A: B=blue area,  R=red area.
$R+B=b\cdot b^{n}-a\cdot a^{n}=\\=b^{n+1}-a^{n+1}$
$R+B=3B+B=4B$
$4B=b^{n+1}-a^{n+1}$
$B=\int_{a}^{b}x^{n}dx=\frac{b^{n+1}}{n+1}-\frac{a^{n+1}}{n+1}$
A: 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
$$\text{Red area}=b^{n+1}-a^{n+1}-\frac{b^{n+1}-a^{n+1}}{n+1}=3\left(\frac{b^{n+1}-a^{n+1}}{n+1}\right)\\\Rightarrow b^{n+1}-a^{n+1}=4\left(\frac{b^{n+1}-a^{n+1}}{n+1}\right)\\\Rightarrow n = 3$$
A: 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!
