IB Maths AA 5.11 Notes

Find the area enclosed by the curve $x=y^2+1$, the $y$-axis, and the lines $y=0$ and $y=2$.

The region runs from $x=0$ to $x=y^2+1$, so $A=\int_0^2 (y^2+1)\,dy$.

Integrate: $A=\left[\frac{y^3}{3}+y\right]_0^2=\frac{8}{3}+2=\frac{14}{3}$.

So the area is $\frac{14}{3}$.

If two curves are written as $x=f(y)$ and $x=g(y)$, then the area between them from $y=a$ to $y=b$ is

$A=\int_a^b (\text{right}-\text{left})\,dy$

Find the area between $x=y+4$ and $x=y^2$ from $y=0$ to $y=2$.

On this interval, $x=y+4$ is to the right of $x=y^2$.

So $A=\int_0^2 ((y+4)-y^2)\,dy$.

Evaluate: $A=\left[\frac{y^2}{2}+4y-\frac{y^3}{3}\right]_0^2=2+8-\frac{8}{3}=\frac{22}{3}$.

So the area is $\frac{22}{3}$.