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2019 Paper 1 Q8
D: 1500.0 B: 1500.0
integration substitution definite integrals algebraic manipulation integration by parts change of variable square root functions

The function \(f\) is defined, for \(x > 1\), by $$f(x) = \int_1^x \sqrt{\frac{t-1}{t+1}} dt.$$ Do not attempt to evaluate this integral.

  1. Show that, for \(x > 2\), $$\int_2^x \sqrt{\frac{u-2}{u+2}} du = 2f\left(\frac{1}{2}x\right).$$
  2. Evaluate in terms of \(f\), for \(x > 0\), $$\int_0^x \sqrt{\frac{u}{u+4}} du.$$
  3. Evaluate in terms of \(f\), for \(x > 5\), $$\int_5^x \sqrt{\frac{u-5}{u+1}} du.$$
  4. Evaluate in terms of \(f\) $$\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du.$$

Solution:

  1. Let \(2t = u\), \begin{align*} \int_2^x \sqrt{\frac{u-2}{u+2}} du &= \int_{t=1}^{t=x/2} \sqrt{\frac{2t-2}{2t+2}}2 \d t \\ &= 2\int_{t=1}^{x/2} \sqrt{\frac{t-1}{t+1}} \d t \\ &= 2f\l\frac{x}{2}\r \end{align*}
  2. Let \(v = u-2\), \begin{align*} \int_0^x \sqrt{\frac{u}{u+4}} du &= \int_{v = 2}^{x+2} \sqrt{\frac{v-2}{v+2}} \d v \\ &= 2 f \l \frac{x+2}{2} \r \end{align*}
  3. Let \(v = u-2, \d v = \d u\) \begin{align*} \int_5^x \frac{u-5}{u+1} du &= \int_3^{x-2} \frac{v-3}{v+3} \d v \\ &= \int_1^{\frac{x-2}{3}} \frac{3t - 3}{3t+3} 3 \d t \\ &= 3 f \l \frac{x-2}{3} \r \end{align*}
  4. Let \(v = u^2, \d v = 2u \d u\)\begin{align*}\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du &= \int_1^2 \sqrt{\frac{u^2}{u^2+4}} u \d u \\ &= \int_1^4 \sqrt{\frac{v}{v+4}} \frac12 \d v \\ &= f \l \frac{4+2}{2} \r - f \l \frac{3}{2} \r \\ &= f(3) - f(\frac32) \end{align*}

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2011 Paper 2 Q1
D: 1600.0 B: 1500.0
curve sketching square root functions graphical solution simultaneous equations algebraic manipulation domain restrictions intersection of curves

  1. Sketch the curve \(y=\sqrt{1-x} + \sqrt{3+x}\;\). Use your sketch to show that only one real value of \(x\) satisfies \[ \sqrt{1-x} + \sqrt{3+x} = x+1\,, \] and give this value.
  2. Determine graphically the number of real values of \(x\) that satisfy \[ 2\sqrt{1-x} = \sqrt{3+x} + \sqrt{3-x}\;. \] Solve this equation.

Solution:

  1. TikZ diagram
    Clearly the only solution is \(x = 1\)
  2. TikZ diagram
    There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \end{align*}

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