Question

Find the area between f(x) = 3x and f(x) = x2 on the interval (0,3]

Answer 1

Answer:

$A=4.5$

Step-by-step explanation:

The area between two curves over the interval $[a,b]$ is $A=\int\limits^b_a {f(x)-g(x)} \, dx$ where $f(x)$ is the upper function and $g(x)$ is the lower function.

Therefore:

$A=\int\limits^3_0 {3x-x^2} \, dx$

$A=\frac{3}{2}x^2-\frac{x^3}{3}\Big|_0^3$

$A=[\frac{3}{2}(3)^2-\frac{(3)^3}{3}]-[\frac{3}{2}(0)^2-\frac{(0)^3}{3}]$

$A=\frac{3}{2}(9)-\frac{27}{3}$

$A=\frac{27}{2}-\frac{27}{3}$

$A=\frac{9}{2}$

$A=4.5$

So, the area between $f(x)=3x$ and $g(x)=x^2$ on the interval $(0,3]$ is $4.5$ square units.