Expanding, splitting fractions and boundary conditions Problem set A Worked solutions

Expand the brackets, then integrate.

1. (a) Find \(\int x^{3}(2x+7)\,dx\).

   • Expand: \(2x^{4} + 7x^{3}\)
   • Integrate: \(\frac{2}{5}x^{5} + \frac{7}{4}x^{4}+c\)

2. (b) Find \(\int (x+5)(x-4)\,dx\).

   • Expand: \(x^{2} + x - 20\)
   • Integrate: \(\frac{1}{3}x^{3} + \frac{1}{2}x^{2} - 20x+c\)

Expand the squared bracket, then integrate.

1. (a) Find \(\int \left(x-3x^{-1}\right)^{2}\,dx\).

   • Expand: \(x^{2} + 9x^{-2} - 6\)
   • Integrate: \(\frac{1}{3}x^{3} - 6x - 9x^{-1}+c\)

2. (b) Find \(\int \left(x^{2}+2x^{-2}\right)^{2}\,dx\).

   • Expand: \(x^{4} + 4x^{-4} + 4\)
   • Integrate: \(\frac{1}{5}x^{5} + 4x - \frac{4}{3}x^{-3}+c\)

Write each fraction as a sum of powers of x, then integrate.

1. (a) Find \(\int \frac{7x^{4}-2x^{2}}{x^{2}}\,dx\).

   • Split into powers of x: \(7x^{2} - 2\)
   • Integrate: \(\frac{7}{3}x^{3} - 2x+c\)

2. (b) Find \(\int \frac{3x+8}{4x^{3}}\,dx\).

   • Split into powers of x: \(\frac{3}{4}x^{-2} + 2x^{-3}\)
   • Integrate: \(-\frac{3}{4}x^{-1} - x^{-2}+c\)

Find \(\int\left(\frac{5}{2t^{3}}-\frac{3}{t^{6}}\right)dt\).

• Write as powers of t: \(\frac{5}{2}t^{-3} - 3t^{-6}\)
• Integrate: \(-\frac{5}{4}t^{-2} + \frac{3}{5}t^{-5}+c\)

Find an expression for y in terms of x.

1. (a) \(\frac{dy}{dx}=8x^{3}-3\) and \(y=10\) when \(x=1\).

   • Integrate: \(y=2x^{4} - 3x+c\)
   • Substitute \(x=1\), \(y=10\): \(10=-1+c\)
   • \(c=11\)
   • \(y=2x^{4} - 3x + 11\)

2. (b) \(\frac{dy}{dx}=9-\frac{4}{x^{2}}\) and \(y=-2\) when \(x=2\).

   • Integrate: \(y=9x + 4x^{-1}+c\)
   • Substitute \(x=2\), \(y=-2\): \(-2=20+c\)
   • \(c=-22\)
   • \(y=9x + 4x^{-1} - 22\)

A curve has gradient \(\frac{dy}{dx}=\frac{10}{x^{2}}-6x\) and passes through the point \((2,3)\). Find the equation of the curve.

• Write as powers: \(-6x + 10x^{-2}\)
• Integrate: \(y=-3x^{2} - 10x^{-1}+c\)
• Substitute \(x=2\), \(y=3\): \(3=-17+c\)
• \(c=20\)
• \(y=-3x^{2} - 10x^{-1} + 20\)

A skateboarder's velocity along a straight path is \(v=\frac{ds}{dt}=3t^{2}-8t+5\) m/s, where s is the displacement in metres from the start and t the time in seconds. At \(t=0\) she is at the start, so \(s=0\).

1. (a) Find an expression for s in terms of t.

   • Integrate: \(s=t^{3} - 4t^{2} + 5t+c\)
   • Put \(t=0\), \(s=0\): \(0=0+c\), so \(c=0\)

2. (b) Find how far she has gone when \(t=3\).

   • Put \(t=3\): \(s=(3)^{3} - 4(3)^{2} + 5(3)=6\)

The gradient of a ski slope is \(\frac{dy}{dx}=2x-\frac{12}{x^{2}}\), where y is the height in metres and x the horizontal distance in hundreds of metres. The slope passes through the point \((3,20)\).

1. (a) Find the equation of the slope.

   • Write as powers: \(2x - 12x^{-2}\)
   • Integrate: \(y=x^{2} + 12x^{-1}+c\)
   • Substitute \(x=3\), \(y=20\): \(20=13+c\)
   • \(c=7\)
   • \(y=x^{2} + 12x^{-1} + 7\)

2. (b) Find the height when \(x=2\).

   • Put \(x=2\): \(y=(2)^{2} + 12(2)^{-1} + 7=17\)

A curve has gradient \(\frac{dy}{dx}=(x+2)(x-5)\) and passes through the point \((3,0)\). Find the equation of the curve.

• Expand: \(x^{2} - 3x - 10\)
• Integrate: \(y=\frac{1}{3}x^{3} - \frac{3}{2}x^{2} - 10x+c\)
• Substitute \(x=3\), \(y=0\): \(0=-\frac{69}{2}+c\)
• \(c=\frac{69}{2}\)
• \(y=\frac{1}{3}x^{3} - \frac{3}{2}x^{2} - 10x + \frac{69}{2}\)

Given that \(\frac{dy}{dx}=\frac{6x^{2}-4}{x^{4}}\) and \(y=2\) when \(x=1\), find y in terms of x.

• Split into powers of x: \(6x^{-2} - 4x^{-4}\)
• Integrate: \(y=-6x^{-1} + \frac{4}{3}x^{-3}+c\)
• Substitute \(x=1\), \(y=2\): \(2=-\frac{14}{3}+c\)
• \(c=\frac{20}{3}\)
• \(y=-6x^{-1} + \frac{4}{3}x^{-3} + \frac{20}{3}\)