Anti-differentiation: reversing the power rule Problem set B Worked solutions

Find f(x) in each case.

1. (a) \(f'(x)=x^{9}\)

   • New power: \(9+1=10\), so divide by 10
   • \(f(x)=\frac{1}{10}x^{10}+c\)

2. (b) \(f'(x)=x^{-5}\)

   • New power: \(-5+1=-4\), so divide by −4
   • \(f(x)=-\frac{1}{4}x^{-4}+c\)

3. (c) \(f'(x)=3\)

   • New power: \(0+1=1\), so divide by 1
   • \(f(x)=3x+c\)

Find y in each case.

1. (a) \(\frac{dy}{dx}=18x^{8}\)

   • Add 1 to each power and divide by the new power: \(18\div 9=2\)
   • \(y=2x^{9}+c\)

2. (b) \(\frac{dy}{dx}=\frac{7}{3}x^{6}\)

   • Add 1 to each power and divide by the new power: \(\frac{7}{3}\div 7=\frac{1}{3}\)
   • \(y=\frac{1}{3}x^{7}+c\)

Find y in each case.

1. (a) \(\frac{dy}{dx}=-6x^{-4}\)

   • Add 1 to each power and divide by the new power: \((-6)\div (-3)=2\)
   • \(y=2x^{-3}+c\)

2. (b) \(\frac{dy}{dx}=\frac{5}{3}x^{-6}\)

   • Add 1 to each power and divide by the new power: \(\frac{5}{3}\div (-5)=-\frac{1}{3}\)
   • \(y=-\frac{1}{3}x^{-5}+c\)

Find y in each case.

1. (a) \(\frac{dy}{dx}=7x^{6} + 8x^{3} - 2\)

   • Add 1 to each power and divide by the new power: \(7\div 7=1\), \(8\div 4=2\), \((-2)\div 1=-2\)
   • \(y=x^{7} + 2x^{4} - 2x+c\)

2. (b) \(\frac{dy}{dx}=\frac{2}{5}x^{4} - \frac{3}{2}x\)

   • Add 1 to each power and divide by the new power: \(\frac{2}{5}\div 5=\frac{2}{25}\), \((-\frac{3}{2})\div 2=-\frac{3}{4}\)
   • \(y=\frac{2}{25}x^{5} - \frac{3}{4}x^{2}+c\)

Find \(\int\left(3x^{2} + 10x^{-6} + 8\right)dx\).

• Add 1 to each power and divide by the new power: \(3\div 3=1\), \(8\div 1=8\), \(10\div (-5)=-2\)
• \(x^{3} + 8x - 2x^{-5} + c\)

Find \(\int\left(\frac{9}{4}y^{-4} + \frac{6}{5}y^{-7}\right)dy\).

• Add 1 to each power and divide by the new power: \(\frac{9}{4}\div (-3)=-\frac{3}{4}\), \(\frac{6}{5}\div (-6)=-\frac{1}{5}\)
• \(-\frac{3}{4}y^{-3} - \frac{1}{5}y^{-6} + c\)

Rewrite as a power of x, then find f(x).

1. (a) \(f'(x)=-\frac{15}{x^{4}}\)

   • Write as a power: \(f'(x)=-15x^{-4}\)
   • Add 1 to each power and divide by the new power: \((-15)\div (-3)=5\)
   • \(f(x)=5x^{-3}+c\)

2. (b) \(f'(x)=\frac{18}{x^{7}}\)

   • Write as a power: \(f'(x)=18x^{-7}\)
   • Add 1 to each power and divide by the new power: \(18\div (-6)=-3\)
   • \(f(x)=-3x^{-6}+c\)

Find f(x) in each case.

1. (a) \(f'(x)=12\sqrt{x}\)

   • Write as a power: \(f'(x)=12x^{1/2}\)
   • Add 1 to each power and divide by the new power: \(12\div \frac{3}{2}=8\)
   • \(f(x)=8x^{3/2}+c\)

2. (b) \(f'(x)=6x^{-1/2}\)

   • Write as a power: \(f'(x)=6x^{-1/2}\)
   • Add 1 to each power and divide by the new power: \(6\div \frac{1}{2}=12\)
   • \(f(x)=12x^{1/2}+c\)

The water in a pond changes at a rate of \(\frac{dV}{dt}=9t^{2}+8t-6\) litres per hour, where V is the volume gained and t the time in hours.

1. (a) Find an expression for V in terms of t, including the constant c.

   • Add 1 to each power and divide by the new power: \(9\div 3=3\), \(8\div 2=4\), \((-6)\div 1=-6\)
   • \(V=3t^{3} + 4t^{2} - 6t+c\)

2. (b) No water has been added when \(t=0\), so \(V=0\). Write down the value of c.

   • Put \(t=0\): \(V=0+c\)
   • \(0=c\), so c = 0

The brightness B of a lamp changes with the distance r metres from it at a rate \(\frac{dB}{dr}=-\frac{15}{r^{4}}+6r\). Find an expression for B in terms of r.

• Write as powers: \(6r - 15r^{-4}\)
• Divide each coefficient by its new power
• \(B=3r^{2} + 5r^{-3}+c\)

Find f(x) when \(f'(x)=\frac{9}{x^{3}}+9\sqrt{x}-8x^{-1/2}\).

• Write every term as a power: \(9x^{1/2} - 8x^{-1/2} + 9x^{-3}\)
• Add 1 to each power and divide by it
• \(f(x)=6x^{3/2} - 16x^{1/2} - \frac{9}{2}x^{-2}+c\)

Priya writes: \(\int 9x^{-3}\,dx=-27x^{-4}+c\). Say what Priya did wrong and give the correct answer.

• Priya multiplied by the power and took 1 off the power. That is differentiating.
• Integrating adds 1 to the power and divides by the new power
• \(\int 9x^{-3}\,dx=-\frac{9}{2}x^{-2}+c\)