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

Find f(x) in each case.

1. (a) \(f'(x)=x^{8}\)

   • New power: \(8+1=9\), so divide by 9
   • \(f(x)=\frac{1}{9}x^{9}+c\)

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

   • New power: \(-6+1=-5\), so divide by −5
   • \(f(x)=-\frac{1}{5}x^{-5}+c\)

3. (c) \(f'(x)=1\)

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

Find y in each case.

1. (a) \(\frac{dy}{dx}=14x^{6}\)

   • Add 1 to each power and divide by the new power: \(14\div 7=2\)
   • \(y=2x^{7}+c\)

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

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

Find y in each case.

1. (a) \(\frac{dy}{dx}=-4x^{-5}\)

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

2. (b) \(\frac{dy}{dx}=\frac{3}{4}x^{-7}\)

   • Add 1 to each power and divide by the new power: \(\frac{3}{4}\div (-6)=-\frac{1}{8}\)
   • \(y=-\frac{1}{8}x^{-6}+c\)

Find y in each case.

1. (a) \(\frac{dy}{dx}=5x^{4} - 6x^{2} + 9\)

   • Add 1 to each power and divide by the new power: \(5\div 5=1\), \((-6)\div 3=-2\), \(9\div 1=9\)
   • \(y=x^{5} - 2x^{3} + 9x+c\)

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

   • Add 1 to each power and divide by the new power: \(\frac{3}{4}\div 3=\frac{1}{4}\), \(\frac{5}{3}\div 2=\frac{5}{6}\)
   • \(y=\frac{1}{4}x^{3} + \frac{5}{6}x^{2}+c\)

Find \(\int\left(4x^{3} + 6x^{-3} - 5\right)dx\).

• Add 1 to each power and divide by the new power: \(4\div 4=1\), \((-5)\div 1=-5\), \(6\div (-2)=-3\)
• \(x^{4} - 5x - 3x^{-2} + c\)

Find \(\int\left(\frac{5}{2}y^{-3} - \frac{4}{3}y^{-5}\right)dy\).

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

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

1. (a) \(f'(x)=-\frac{10}{x^{3}}\)

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

2. (b) \(f'(x)=\frac{12}{x^{5}}\)

   • Write as a power: \(f'(x)=12x^{-5}\)
   • Add 1 to each power and divide by the new power: \(12\div (-4)=-3\)
   • \(f(x)=-3x^{-4}+c\)

Find f(x) in each case.

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

   • Write as a power: \(f'(x)=9x^{1/2}\)
   • Add 1 to each power and divide by the new power: \(9\div \frac{3}{2}=6\)
   • \(f(x)=6x^{3/2}+c\)

2. (b) \(f'(x)=10x^{-1/2}\)

   • Write as a power: \(f'(x)=10x^{-1/2}\)
   • Add 1 to each power and divide by the new power: \(10\div \frac{1}{2}=20\)
   • \(f(x)=20x^{1/2}+c\)

The velocity of a cyclist along a straight track is \(v=\frac{ds}{dt}=6t^{2}-10t+4\) m/s, where s is the displacement in metres from the start line and t the time in seconds.

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

   • Add 1 to each power and divide by the new power: \(6\div 3=2\), \((-10)\div 2=-5\), \(4\div 1=4\)
   • \(s=2t^{3} - 5t^{2} + 4t+c\)

2. (b) The cyclist is at the start line when \(t=0\), so \(s=0\). Write down the value of c.

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

The strength S of a radio signal changes with the distance d km from the transmitter at a rate \(\frac{dS}{dd}=-\frac{12}{d^{3}}+2d\). Find an expression for S in terms of d.

• Write as powers: \(2d - 12d^{-3}\)
• Divide each coefficient by its new power
• \(S=d^{2} + 6d^{-2}+c\)

Find f(x) when \(f'(x)=\frac{4}{x^{2}}+6\sqrt{x}-3x^{-1/2}\).

• Write every term as a power: \(6x^{1/2} - 3x^{-1/2} + 4x^{-2}\)
• Add 1 to each power and divide by it
• \(f(x)=4x^{3/2} - 6x^{1/2} - 4x^{-1}+c\)

Sam writes: \(\int 6x^{-4}\,dx=-24x^{-5}+c\). Say what Sam did wrong and give the correct answer.

• Sam 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
• \(\int 6x^{-4}\,dx=-2x^{-3}+c\)