Finding an Unknown Integration Limit | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Turn the required area into an equation

When a boundary is unknown, keep it as a letter while evaluating the integral. Equate the result to the stated area, then solve. Half an area does not usually mean half the width: the strip heights can vary.

Worked example

The line x = p halves the area under y = 12/x² between x = 1 and x = 4. Find p.

The function is positive throughout the region. Its total area is:

\begin{gathered}\int_1^4 12x^{-2}\,dx\\=\left[-\frac{12}{x}\right]_1^4\\=-3+12\\=9\end{gathered}

Require the part from 1 to $p$ to have area $9/2$ , with $1<p<4$ .

\begin{aligned}\int_1^p12x^{-2}\,dx&=\frac92\\12-\frac{12}{p}&=\frac92\\\frac{12}{p}&=\frac{15}{2}\\p&=\frac85\end{aligned}

The graph is taller near 1, so half the area occupies less than half the width. The value $p=1.6$ is consistent with that check.

Reject limits outside the stated region

Your turnPractice

The line $x=p$ halves the region under $y=2x$ from 0 to 6. Find $p$ exactly.

Show worked answer

The total area is 36, so $\int_0^p2x\,dx=p^2=18$ .

Answer

p=3\sqrt2

The other algebraic root is negative and fails $0<p<6$ . The increasing height means more than half the width is needed to reach half the total area.

Your turnUse given integrals

Given $\int_1^5 f(x)\,dx=9$ and $\int_1^2 f(x)\,dx=4$ , find $\int_5^2 2f(x)\,dx$ without knowing $f(x)$ .

Show worked answer

The integral from 2 to 5 is $9-4=5$ . Reverse the limits and multiply by two.

Answer

-10

Turn the required area into an equation

Worked example

The line x = p halves the area under y = 12/x² between x = 1 and x = 4. Find p.

The function is positive throughout the region. Its total area is:

\begin{gathered}\int_1^4 12x^{-2}\,dx\\=\left[-\frac{12}{x}\right]_1^4\\=-3+12\\=9\end{gathered}

Require the part from 1 to $p$ to have area $9/2$ , with $1<p<4$ .

\begin{aligned}\int_1^p12x^{-2}\,dx&=\frac92\\12-\frac{12}{p}&=\frac92\\\frac{12}{p}&=\frac{15}{2}\\p&=\frac85\end{aligned}

The graph is taller near 1, so half the area occupies less than half the width. The value $p=1.6$ is consistent with that check.

Reject limits outside the stated region

Your turnPractice

The line $x=p$ halves the region under $y=2x$ from 0 to 6. Find $p$ exactly.

Show worked answer

The total area is 36, so $\int_0^p2x\,dx=p^2=18$ .

Answer

p=3\sqrt2

The other algebraic root is negative and fails $0<p<6$ . The increasing height means more than half the width is needed to reach half the total area.

Your turnUse given integrals

Given $\int_1^5 f(x)\,dx=9$ and $\int_1^2 f(x)\,dx=4$ , find $\int_5^2 2f(x)\,dx$ without knowing $f(x)$ .

Show worked answer

The integral from 2 to 5 is $9-4=5$ . Reverse the limits and multiply by two.

Answer

-10