Question

A right triangle has one vertex on the graph of $y=x^3,x>0,$at $\left(x,y\right)$, another at the origin, and the third on the positive y-axis at $\left(0,y\right)$, as shown in the figure. Express the area Aof the triangle as a function of x.

Short answer

The area of the given triangles as function is $A\left(x\right)=\frac{1}{2}{x}^4$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

One vertex is at $\left(x,y\right)$.

Second vertex is at $\left(0,0\right)$.

Third vertex is at $\left(0,y\right)$.

We know the area of the triangle is $A=\frac{1}{2}\times base\times height$.

Step 2. Substitute the values.

Consider the given figure,

$A=\frac{1}{2}\times x\times y......\left(i\right)$

As $\left(x,y\right)$is a point on the graph of the function $y=x^3$.

Substitute it in equation (i),

role="math" localid="1645786718954" $$\begin{gathered} A\left(x\right)=\frac{1}{2}\times x\times {\left(x\right)}^3 \\ A\left(x\right)=\frac{1}{2}{x}^4 \end{gathered}$$