Final answer:
To find the enclosed area by the equations y=8x and x=3, calculate the definite integral of 8x from x=0 to x=3, which results in 36 square units.
Step-by-step explanation:
To find the area enclosed by the graphs of the equations y=8x, and x=3, you can set up an integral with respect to the x-axis.
The graph of y=8x is a straight line passing through the origin with a slope of 8. The graph of x=3 is a vertical line that intersects the x-axis at x=3.
These two graphs intersect at the point (3, 24), because when x=3, y=8(3)=24.
We want to find the area of the region from x=0 to x=3 under the curve y=8x.
To do this, we calculate the definite integral of the function f(x)=8x from x=0 to x=3:
int_0^3 8x dx
You evaluate the integral using the basic integral rule:
int x^n dx = {1} / {n+1}x^{n+1}+C
Applying this to our integral:
int_0^3 8x dx = 4x^2 Big|_0^3
= 4(3)^2 - 4(0)^2
= 4(9) - 0
= 36
Thus, the enclosed area is 36 square units.