Integrating with Respect to y | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Use horizontal strips when the limits are heights

A region bounded by the y-axis and two horizontal lines is often simpler to measure horizontally. Each thin strip has width right x-coordinate minus left x-coordinate, and small height along $y$ . Add these strips by integrating with respect to $y$ , written $dy$ .

A horizontal strip extends from x = 0 to x = (y² − 1)/2. Its thickness is measured along y.

Use y-values for the limits and express the width entirely in $y$ . The power rule is unchanged: for example, $\int y^2\,dy=y^3/3+c$ .

Express the boundary in the integration variable

Worked example

Find the area bounded by y = √(2x + 1), the y-axis and y = 3.

The curve meets the y-axis at $y=1$ . Squaring the given relation and rearranging gives $x=(y^2-1)/2$ . For $1\le y\le3$ , the left boundary is $x=0$ .

\begin{aligned}A&=\int_1^3\frac{y^2-1}{2}\,dy\\&=\left[\frac{y^3}{6}-\frac y2\right]_1^3\\&=3-\left(-\frac13\right)=\frac{10}{3}\end{aligned}

The region also equals the $4\times3$ rectangle minus the area under the curve from $x=0$ to 4. Both methods measure the same shaded region.

Your turnPractice

Find the area bounded by $x=y(4-y)$ and the y-axis. Why is y a convenient integration variable?

Show worked answer

The intersections with $x=0$ are $y=0,4$ . At each height between them, one right-hand x-value is already given.

\begin{gathered}A=\int_0^4(4y-y^2)\,dy\\=\left[2y^2-\frac{y^3}{3}\right]_0^4\\=\frac{32}{3}\end{gathered}

Your turnUse the region

The area under $y=x^2$ from 0 to 2 is $8/3$ . Find the area between this curve, the y-axis and $y=4$ without another integration.

Show worked answer

The two regions fill a rectangle of width 2 and height 4.

Answer

A=8-\frac83=\frac{16}{3}

Use horizontal strips when the limits are heights

A horizontal strip extends from x = 0 to x = (y² − 1)/2. Its thickness is measured along y.

Use y-values for the limits and express the width entirely in $y$ . The power rule is unchanged: for example, $\int y^2\,dy=y^3/3+c$ .

Express the boundary in the integration variable

Worked example

Find the area bounded by y = √(2x + 1), the y-axis and y = 3.

The curve meets the y-axis at $y=1$ . Squaring the given relation and rearranging gives $x=(y^2-1)/2$ . For $1\le y\le3$ , the left boundary is $x=0$ .

\begin{aligned}A&=\int_1^3\frac{y^2-1}{2}\,dy\\&=\left[\frac{y^3}{6}-\frac y2\right]_1^3\\&=3-\left(-\frac13\right)=\frac{10}{3}\end{aligned}

The region also equals the $4\times3$ rectangle minus the area under the curve from $x=0$ to 4. Both methods measure the same shaded region.

Your turnPractice

Find the area bounded by $x=y(4-y)$ and the y-axis. Why is y a convenient integration variable?

Show worked answer

The intersections with $x=0$ are $y=0,4$ . At each height between them, one right-hand x-value is already given.

\begin{gathered}A=\int_0^4(4y-y^2)\,dy\\=\left[2y^2-\frac{y^3}{3}\right]_0^4\\=\frac{32}{3}\end{gathered}

Your turnUse the region

The area under $y=x^2$ from 0 to 2 is $8/3$ . Find the area between this curve, the y-axis and $y=4$ without another integration.

Show worked answer

The two regions fill a rectangle of width 2 and height 4.

Answer

A=8-\frac83=\frac{16}{3}