Area Between a Curve & y-Axis (College Board AP® Calculus AB): Revision Note

In an FRQ, make sure you clearly state the integrand and the limits of integration. Make sure the variable in the differential, matches the variable in the integrand.

Find the area of the region enclosed by the curve with equation $y=2+\sqrt{x+4}$, the $y$-axis, and the horizontal line $y=6$. The graph of the curve is shown below.

Answer:

The curve crosses the $y$-axis at (0, 4)

The diagram shows that the integral for the area between the curve, the $y$-axis, and $y=6$ will be only positive (none of the area is to the left of the $y$-axis)

Rearrange the equation for $y$ in terms of $x$, to make an equation for $x$ in terms of $y$

$\begin{array}{rcl}y& =& 2+\sqrt{x+4}\\ y-2& =& \sqrt{x+4}\\ {\left(y-2\right)}^2& =& x+4\\ {\left(y-2\right)}^2-4& =& x\end{array}$

Integrate this with respect to $y$ between the $y$ values of 4 and 6

$\begin{array}{rcl}{\int }_4^6{\left(y-2\right)}^2-4\mathit{d}y& =& {\left[\frac{1}{3}{\left(y-2\right)}^3-4y\right]}_4^6\\ & =& \left(\frac{1}{3}{\left(6-2\right)}^3-4\left(6\right)\right)-\left(\frac{1}{3}{\left(4-2\right)}^3-4\left(4\right)\right)\\ & =& \left(-\frac{8}{3}\right)-\left(-\frac{40}{3}\right)\\ & =& \frac{32}{3}\end{array}$

$\frac{32}{3}$ units squared

Always check whether you need to find the value of an integral, or an area.

• When areas to the left of the $y$-axis are involved, these will be two different values.

Find the area of the region enclosed by the curve with equation $x=y^2-7y+10$ and the $y$-axis.

Answer:

Notice that in this question we are already given an equation for $x$ in terms of $y$

We are not told any limits for this question, so we need to find where the $y$-intercepts are by solving for $x=0$

$$\begin{gathered} 0=y^2-7y+10 \\ 0=\left(y-2\right)\left(y-5\right) \\ y=2\mathrm{or}y=5 \end{gathered}$$

The $y$-intercepts are at $y=2$ and $y=5$, this can be used to sketch a graph

It can now be seen that the area enclosed by the curve and the $y$-axis will be entirely on the left of the axis, and so the value of the integral will be negative

Remember to make this positive at the end, as it asks for an area

Integrate with respect to $y$ between the $y$ values of 2 and 5

$\begin{array}{rcl}{\int }_2^5{y}^2-7y+10\mathit{d}y& =& {\left[\frac{1}{3}{y}^3-\frac{7}{2}{y}^2+10y\right]}_2^5\\ & =& \left(\frac{1}{3}\left(5^3\right)-\frac{7}{2}\left(5^2\right)+10\left(5\right)\right)-\left(\frac{1}{3}\left(2^3\right)-\frac{7}{2}\left(2^2\right)+10\left(2\right)\right)\\ & =& \frac{25}{6}-\frac{26}{3}\\ & =& -\frac{9}{2}\end{array}$

The question requires an area, rather than the value of the integral, so make this positive

$\frac{9}{2}$ units squared