The area enclosed by the curves y=x^2, y=x^3, x=0 and x=p, where p1, is 16.

Question

The area enclosed by the curves $y=x^2$, $y=x^3$, $x=0$ and $x=p$, where $p>1$, is $\dfrac16$. Then $p$ equals:

Accepted Answer

$x^2=x^3$ at $x=1$.

$$\frac16 = \int_0^1(x^2-x^3)\,dx + \int_1^p(x^3-x^2)\,dx = \left(\frac13-\frac14\right) + \left(\frac{p^4}{4}-\frac{p^3}{3}\right) - \left(\frac14-\frac13\right)$$

$$\frac16 = \frac16 + \frac{p^4}{4}-\frac{p^3}{3} \implies \frac{p^4}{4}=\frac{p^3}{3} \implies p = \frac43$$

Solutions · 1