Respuesta :
Answer:
[tex]\frac{a^3}{6}[/tex]
Step-by-step explanation:
First let's find the [tex]x[/tex]'s for which the curves form a enclosure.
We do this by finding when f and g intersect.
[tex]ax=x^2[/tex]
Subtract [tex]ax[/tex] on both sides:
[tex]0=x^2-ax[/tex]
Factor right hand side:
[tex]0=x(x-a)[/tex]
The product is zero when at least one of it's factors is zero.
So [tex]x=0[/tex] or [tex]x=a[/tex]
[tex]a[/tex] is positive and is greater than 0 so [tex]a[/tex] is the upper bound.
Also [tex]ax>x^2[/tex] when [tex]x[/tex] is between [tex]0 \text{and} a[/tex].
The integral we need to evaluate is:
[tex]\int_0^a (ax-x^2)dx[/tex]
[tex](\frac{ax^2}{2}-\frac{x^3}{3})_0^a[/tex]
Plug in limits:
[tex](\frac{aa^2}{2}-\frac{a^3}{3})- (\frac{a0^2}{2}-\frac{0^3}{3})[/tex]
[tex]\frac{a^3}{2}-\frac{a^3}{3}[/tex]
[tex]\frac{3a^3-2a^3}{6}[/tex]
[tex]\frac{a^3}{6}[/tex]
