Integrating Powers of a Linear Bracket | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Reverse both factors from the chain rule

A linear bracket has the form $ax+b$ , where $a$ and $b$ are constants. Its derivative is the constant $a$ . You can keep this bracket intact instead of expanding a high power.

\frac{d}{dx}(2x-3)^6=6(2x-3)^5\times2

Integration must undo the outer factor 6 and the inner factor 2. In general, increase the power by one and divide by both the new power and the inner gradient:

\begin{gathered}\int(ax+b)^n\,dx\\=\frac{(ax+b)^{n+1}}{a(n+1)}+c\end{gathered}

This requires $a\ne0$ and $n\ne-1$ . If $a=0$ , the integrand is just a constant where defined. Roots and denominators must also remain defined on the interval used.

Keep the bracket and its sign unchanged

Worked example

Integrate 12(2x − 3)⁵.

\begin{gathered}\int12(2x-3)^5\,dx\\=\frac{12}{2\times6}(2x-3)^6+c\end{gathered}

Answer

(2x-3)^6+c

Differentiating produces $6\times2=12$ , as needed.

The same correction works with negative and fractional powers. For $x<5/2$ , rewrite the reciprocal square root first:

\begin{gathered}\int\frac3{\sqrt{5-2x}}\,dx\\=\frac3{(-2)(1/2)}(5-2x)^{1/2}+c\\=-3\sqrt{5-2x}+c\end{gathered}

The negative inner coefficient matters. Differentiating this answer gives a positive integrand.

Your turn9709/12 Oct/Nov 2021 Q4 [4 marks]

A curve has $dy/dx=8(3x+2)^{-2}$ and passes through $(2,17/3)$ . Find its equation.

Show worked answer

\begin{gathered}y=-\frac8{3(3x+2)}+c\\\frac{17}{3}=-\frac13+c\\c=6\end{gathered}

Answer

y=6-\frac8{3(3x+2)}

The curve containing the given point lies on the side $x>-2/3$ of the excluded denominator zero.

Reverse both factors from the chain rule

A linear bracket has the form $ax+b$ , where $a$ and $b$ are constants. Its derivative is the constant $a$ . You can keep this bracket intact instead of expanding a high power.

\frac{d}{dx}(2x-3)^6=6(2x-3)^5\times2

Integration must undo the outer factor 6 and the inner factor 2. In general, increase the power by one and divide by both the new power and the inner gradient:

\begin{gathered}\int(ax+b)^n\,dx\\=\frac{(ax+b)^{n+1}}{a(n+1)}+c\end{gathered}

This requires $a\ne0$ and $n\ne-1$ . If $a=0$ , the integrand is just a constant where defined. Roots and denominators must also remain defined on the interval used.

Keep the bracket and its sign unchanged

Worked example

Integrate 12(2x − 3)⁵.

\begin{gathered}\int12(2x-3)^5\,dx\\=\frac{12}{2\times6}(2x-3)^6+c\end{gathered}

Answer

(2x-3)^6+c

Differentiating produces $6\times2=12$ , as needed.

The same correction works with negative and fractional powers. For $x<5/2$ , rewrite the reciprocal square root first:

\begin{gathered}\int\frac3{\sqrt{5-2x}}\,dx\\=\frac3{(-2)(1/2)}(5-2x)^{1/2}+c\\=-3\sqrt{5-2x}+c\end{gathered}

The negative inner coefficient matters. Differentiating this answer gives a positive integrand.

Your turn9709/12 Oct/Nov 2021 Q4 [4 marks]

A curve has $dy/dx=8(3x+2)^{-2}$ and passes through $(2,17/3)$ . Find its equation.

Show worked answer

\begin{gathered}y=-\frac8{3(3x+2)}+c\\\frac{17}{3}=-\frac13+c\\c=6\end{gathered}

Answer

y=6-\frac8{3(3x+2)}

The curve containing the given point lies on the side $x>-2/3$ of the excluded denominator zero.