Chapter 2.6, Problem 5AYU

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

Express the area A of the triangle as a function of x.

To determine

To find: To express the area $A$ of the triangle as a function of $x$.

Answer to Problem 5AYU

Solution:

$\frac{1}{2}{x}^4$

Explanation of Solution

Given:

$y=x^3, x>0$

Calculation:

To express the area $A$ of the triangle as a function of $x$:

Area of the given triangle $=\frac{1}{2}xy$

Since the given right triangle has one vertex on the graph $y=x^3, x>0$.

Substitute $x^3$ for $y$. Then

Area of the triangle $=\frac{1}{2}x x^3$

$=\frac{1}{2}{x}^4$