Find $n$ if the area between the curve of $y=x^n$ and the $y$ axis is $3$ times the area between the curve and the $x$-axis 
For this question i tried to find the area for red and blue sections, and equate them by red= 3 blue. 
However, it didnt work out and I got $b-a = 3(b^n - a^n)$ for the outcome. 
In the book, it says we dont have to deal with the red area....
Thank you.
 A: The blue area is $$\int_a^b x^n dx = \frac{b^{n+1}-a^{n+1}}{n+1}.$$
The red (orange?) area is the area of the big rectangle with sides, $b$, $b^n$, minus the small (white) rectangle with sides $a$, $a^n$, minus the above integral. That is $$b^{n+1}-a^{n+1}-\frac{b^{n+1}-a^{n+1}}{n+1}.$$ Hence, we need $$ b^{n+1}-a^{n+1}-\frac{b^{n+1}-a^{n+1}}{n+1} = 3 \cdot \frac{b^{n+1}-a^{n+1}}{n+1}. $$ Simplifying a bit gives $$ (n+1)(b^{n+1} - a^{n+1}) = 4 (b^{n+1} - a^{n+1}). $$ So we have $$(n - 3)(b^{n+1}-a^{n+1})=0.$$ Since $a$ and $b$ are arbitrary, we require $n=3$.
A: HINT
Blue area is area under $f(x) = x^n$, so $$B = \int_a^b x^n dx$$ and red area is $$R = \int_a^b x^{1/n} dx$$ and you need $$R = 3B$$
But a different way is noticing the rectangle of sides $b, b^n$ is made up of red, blue and a rectangle of sides $a,a^n$ so you get
$$
R = b \cdot b^n - a \cdot a^n - B...
$$
A: The book is correct.  The area of the rectangular area is $b^{n+1}-a^{n+1}$ while we are given that the areas of the red and blue sections add to $4$ times the area of the blue section.  The area of the blue section is
$$\int_a^b x^n\,dx=\frac{b^{n+1}-a^{n+1}}{n+1}$$
Therefore, we have 
$$4\frac{b^{n+1}-a^{n+1}}{n+1}=b^{n+1}-a^{n+1}\implies n=3$$
A: First, the question only has one answer if you are looking at the unit square as the encompassing rectangle: Solving $$\frac{1-X}{X}=3$$ where $X$ is the area between the curve and the $y$-axis, you need $X=\frac{1}{4}$. Since $$\int_0^1x^n\,dn=\frac{1}{n+1}$$ $n$ must be $3$.
Now if you let $b$ be something else besides $1$, you have $$A=\int_0^bx^3\,dx=\frac14b^4$$ and $$B=b\cdot b^3-A =\frac34b^4$$ with $B=3A$.
