AI HL · 13.5
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Expanding, splitting fractions and boundary conditions
13.5.3
Find \(\int (x+6)(x-2)\,dx\).
Have a go first. Then open the working one line at a time.
\((x+6)(x-2)=x^{2}+4x-12\)
The rule works on terms \(ax^{n}\), not on a product of brackets.
\(\frac{x^{3}}{3}+\frac{4x^{2}}{2}-12x\)
Add 1 to each power and divide by it.
\(\frac13x^{3}+2x^{2}-12x+c\)
\(\frac{4}{2}=2\).
Find \(\int \left(x^{2}-2x^{-1}\right)^{2}dx\).
Have a go first. Then open the working one line at a time.
\(x^{4}-2(x^{2})(2x^{-1})+4x^{-2}=x^{4}-4x+4x^{-2}\)
Square the first term, double the product, square the last. \(x^{2}\times x^{-1}=x\).
\(\frac{x^{5}}{5}-\frac{4x^{2}}{2}+\frac{4x^{-1}}{-1}\)
New powers 5, 2 and −1.
\(\frac15x^{5}-2x^{2}-4x^{-1}+c\)
\(\frac{4}{-1}=-4\).
Find \(\int \frac{4x+9}{x^{4}}\,dx\).
Have a go first. Then open the working one line at a time.
\(\frac{4x}{x^{4}}+\frac{9}{x^{4}}\)
Divide each term of the top by the bottom.
\(4x^{1-4}+9x^{-4}=4x^{-3}+9x^{-4}\)
\(\frac{x^{m}}{x^{n}}=x^{m-n}\) and \(\frac{1}{x^{4}}=x^{-4}\).
\(\frac{4}{-2}x^{-2}+\frac{9}{-3}x^{-3}+c=-2x^{-2}-3x^{-3}+c\)
New powers −2 and −3.
Find \(\int\left(\frac{2}{3t^{2}}-\frac{6}{t^{4}}\right)dt\).
Have a go first. Then open the working one line at a time.
\(\frac23t^{-2}-6t^{-4}\)
\(\frac{2}{3t^{2}}=\frac23t^{-2}\). Only the t is in the denominator.
\(\frac{2/3}{-1}t^{-1}-\frac{6}{-3}t^{-3}\)
New powers −1 and −3.
\(-\frac23t^{-1}+2t^{-3}+c\)
The variable is t, so the answer is in t.
If \(\frac{dy}{dx}=6x^{2}-4\) and \(y=7\) when \(x=2\), find an expression for y in terms of x.
Have a go first. Then open the working one line at a time.
\(y=2x^{3}-4x+c\)
Keep c for now.
\(7=2(2)^{3}-4(2)+c=8+c\)
The boundary condition gives one equation in c.
\(c=7-8=-1\)
Move the 8 across.
\(y=2x^{3}-4x-1\)
Write the full expression for y.
A curve has gradient \(\frac{dy}{dx}=\frac{6}{x^{2}}+2x\) and passes through \((3,10)\). Find the equation of the curve.
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=6x^{-2}+2x\)
\(\frac{6}{x^{2}}=6x^{-2}\).
\(y=-6x^{-1}+x^{2}+c\)
\(\frac{6}{-1}=-6\).
\(10=-6(3)^{-1}+3^{2}+c=-2+9+c\)
The point \((3,10)\) is the boundary condition: \(x=3\), \(y=10\).
\(c=3,\quad y=x^{2}-6x^{-1}+3\)
Check by differentiating: \(2x+6x^{-2}\).