A calculus problem by Dexter Woo Teng Koon

I honestly couldn't find an exact answer, so I did a numerical one:

We know that the whole area is:

$\large \displaystyle A = \int_0^1 (x^3 - x^4)dx$

$\color{#20A900} \boxed{\large \displaystyle A = \frac{1}{20}}$

If we call the intersection points of $y = x^3 - x^4$ with $y = h$ as $a$ and $b$ , the area of the curve above $h$ between $a$ and $b$ must be half of $A$ . Hence:

$\large \displaystyle \int_a^b (x^3 - x^4 - h) dx = \frac{1}{40}$

$\color{#20A900} (i) \ \boxed{ \large \displaystyle \frac{1}{4} (b^4 - a^4) - \frac{1}{5} (b^5 - a^5) - h(b-a) - \frac{1}{40} = 0 }$

Also, since $a$ and $b$ are intersection points:

$\color{#20A900} (ii) \ \boxed{ \large \displaystyle a^3 - a^4 - h = 0}$

$\color{#20A900} (iii) \ \boxed{ \large \displaystyle b^3 - b^4 - h = 0}$

Putting $(i)$ , $(ii)$ and $(iii)$ in the Newton's method :

$\large \displaystyle \begin{bmatrix} a_{k+1} \\ b_{k+1} \\ h_{k+1} \end{bmatrix} = \begin{bmatrix} a_{k} \\ b_{k} \\ h_{k} \end{bmatrix} - \begin{bmatrix} -a_k^3 + a_k^4 + h & b_k^3 - b_k^4 - h & a_k - b_k \\ 3a_k^2 - 4a_k^3 & 0 & -1 \\ 0 & 3b_k^2 - 4b_k^3 & - 1 \end{bmatrix} \cdot \begin{bmatrix} \frac{1}{4} (b_k^4 - a_k^4) - \frac{1}{5} (b_k^5 - a_k^5) - h_k(b_k-a_k) - \frac{1}{40} \\ a_k^3 - a_k^4 - h_k \\ b_k^3 - b_k^4 - h_k \end{bmatrix}$

Beginning from $a_0 = 0.3$ , $b_0 = 0.8$ and $h_0 = 0.05$ , it quickly converges to:

$\color{#20A900} \boxed{ \large \displaystyle \begin{bmatrix} a \\ b \\ h \end{bmatrix} = \begin{bmatrix} 0.387256572 \\ 0.95974631 \\ 0.035585662 \end{bmatrix} }$

So:

$\color{#3D99F6} \boxed{ \large \displaystyle h = 0.035585662}$