find the area between the curves f(x)=x^2 and g(x)=ax (assume a>0)

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]