Animated Solution for Mathematics - Definite Integration: It the area enclosed by the parabolas $P_1:2y=5x^2$ and $P_2:x^2-y+6=0$ is equal to the area enclosed by $P_1$ and $y=\alpha x,\alpha >0$, then ${\alpha }^3$ is equal to ______.

Imagine you are standing on a coordinate plane, looking at two paths carved by the equations $P_1:2y=5x^2$ and $P_2:x^2-y+6=0$. These aren't just abstract curves; they are boundaries defining a hidden, enclosed world.

To measure the area of this world, we first rewrite our equations in the standard form $y=f(x)$. We have $P_1:y=\frac{5}{2}{x}^2$ and $P_2:y=x^2+6$.

Notice the beauty here: $P_1$ is a steep, narrow parabola anchored at the origin, while $P_2$ is a broader parabola that has been lifted $6$ units into the air.

Finding the Intersection

To find the area trapped between these two, we must first find where they meet. We equate their $y$-values:

With a bit of algebraic grace, we subtract $x^2$ from both sides to get $\frac{3}{2}{x}^2=6$. Multiplying by $\frac{2}{3}$, we find $x^2=4$, which gives us the intersection points $x=\pm 2$.

These are our boundaries. We are integrating from $-2$ to $2$.

Now, we set up the integral for the area $A_1={\int }_{-2}^2(y_{upper}-y_{lower})dx$. Looking at our curves, $P_2$ is the upper boundary.

So, the integral becomes:

Here is where we can be clever. The function $6-\frac{3}{2}{x}^2$ is an even function, meaning it is symmetric about the $y$-axis. Instead of slogging through the full interval, we can integrate from $0$ to $2$ and double the result:

Performing the integration, we get $2{\left[6x-\frac{1}{2}{x}^3\right]}_0^2$. Evaluating this at the limits, we find $2(12-4)=16$. The area of our first region is exactly $16$ square units.

The Line and the Parabola

Now, the problem shifts. We are looking for a new area $A_2$ enclosed by $P_1:y=\frac{5}{2}{x}^2$ and a line $y=\alpha x$.

Again, we find the intersection by setting $\frac{5}{2}{x}^2=\alpha x$. Factoring out $x$, we get $x(\frac{5}{2}x-\alpha )=0$. The intersection points are $x=0$ and $x=\frac{2\alpha }{5}$.

The area $A_2$ is the integral of the line minus the parabola:

Integrating term by term, we get:

Substituting the upper limit, we obtain:

Finding a common denominator of $75$, we get $\frac{6{\alpha }^3-4{\alpha }^3}{75}=\frac{2{\alpha }^3}{75}$.

We are told that $A_1=A_2$. So, we set $16=\frac{2{\alpha }^3}{75}$.

Solving for ${\alpha }^3$, we multiply $16$ by $75$ and divide by $2$:

Through the power of integration and the symmetry of parabolas, we have arrived at our destination. The value of ${\alpha }^3$ is $600$.