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2025 Paper 2 Q5
D: 1500.0 B: 1500.0
integration substitution improper integrals symmetry generalisation reciprocal substitution algebraic manipulation

You need not consider the convergence of the improper integrals in this question.

  1. Use the substitution \(x = u^{-1}\) to show that \[\int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx = 0.\]
  2. Use the substitution \(x = u^{-2}\) to show that \[\int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx = 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, dx.\]
  3. Find, in terms of \(p\) and \(s\), a value of \(r\) for which \[\int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx = 0,\] given that \(p\) and \(s\) are fixed values for which the required integrals converge.
  4. Show that, for any positive value of \(k\), it is possible to find values of \(p\) and \(q\) for which \[\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx = k\int_0^{\infty} \frac{1}{\sqrt{x^q+1}} \, dx.\]

Solution:

  1. \begin{align*} && I &= \int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx \\ x = u^{-1}, \d x = -u^{-2} \d u: && &= \int_{u=\infty}^{u = 0} \frac{u^{-1/2} - 1}{\sqrt{u^{-1}(u^{-3}+1)}} (-u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-1/2} -1}{\sqrt{1+u^3}} \d u \\ &&&= \int_0^\infty \frac{1-\sqrt{u}}{\sqrt{u(u^3+1)}} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
  2. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx \\ x = u^{-2}, \d x = -2u^{-3} \d u: && &= \int_{u = \infty}^{u = 0} \frac{1}{\sqrt{u^{-6}+1}} (-2u^{-3}) \d u \\ &&&= \int_0^\infty \frac{2}{\sqrt{1+u^6}} \d u \\ &&&= 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, \d x \end{align*}
  3. \begin{align*} && I &= \int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx \\ x = u^{-1}, \d x = - u^{-2} \d t: &&&= \int_{u=\infty}^{u = 0} \frac{u^{-r}-1}{\sqrt{u^{-s}(u^{-p}+1)}} (- u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-r}-1}{\sqrt{u^{4-s}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r-p}(1+u^p)}} \d u \\ \end{align*} Therefore if \(4-s+2r-p = s\) or \(r = \frac{p}2+s-2\) we have \(I = -I\), ie \(I = 0\).
  4. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx \\ x = u^{-t}, \d x = -t u^{-t-1}:&&&= \int_{u = -\infty}^{u = 0} \frac{1}{\sqrt{u^{-pt}+1}} (-t u^{-t-1}) \d u \\ &&&= t \int_0^\infty \frac1{\sqrt{u^{-pt+2t+2}+u^{2t+2}}} \d u \end{align*} Therefore if \(\begin{cases} 2t+2 &= q \\ 2t+2 -pt &= 0 \end{cases}\), ie \(q = 2t+2, p = 2 + 2/t\) we will have found the \(p, q\) desired for any \(t\) (or \(k\)). [Alternatively] Let \(\displaystyle I(p) =\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx\), then clearly \(I(p)\) is decreasing and as \(p \to \infty\) \(I(p) \to 1\), so our integral can take any values on \((1, \infty)\) and so for any positive value we can find two values with a given ratio. In particular given \(k\) and \(p\) we can find a suitable \(q\).

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2000 Paper 3 Q2
D: 1700.0 B: 1484.2
integration trigonometric substitution definite integral square root generalisation trigonometric identities

Use the substitution \(x = 2-\cos \theta \) to evaluate the integral $$ \int_{3/2}^2 \left(x - 1 \over 3 - x\right)^{\!\frac12}\! \d x. $$ Show that, for \(a < b\), $$ \int_p^q \left( x - a \over b - x\right)^{\!\frac12} \!\d x = \frac{(b-a)(\pi +3{\surd3} -6)}{12}, $$ where \(p= {(3a+b)/4}\) and \(q={(a+b)/2}\).

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1993 Paper 3 Q9
D: 1700.0 B: 1485.7
inequality proof AM-GM inequality sum of squares generalisation algebraic manipulation quadratic forms

For the real numbers \(a_1\), \(a_2\), \(a_3\), \(\ldots\),

  1. prove that \(a_1^2+a_2^2 \ge 2a_1a_2\),
  2. prove that \(a_1^2+a_2^2 +a_3^2 \ge a_2a_3 + a_3a_1 +a_1a_2\),
  3. prove that $3(a_1^2+a_2^2 +a_3^2 +a_4^2) \ge 2(a_1a_2+a_1a_3 + a_1a_4 +a_2a_3 + a_2a_4 +a_3a_4)\(,
  4. state and prove a generalisation of (iii) to the case of \)n$ real numbers,
  5. prove that $$ \left(\sum_{i=1}^n a_i \right)^2 \ge {2n\over n-1} \sum_{i,j} a_ia_j, $$ where the latter sum is taken over all pairs \((i,j)\) with $1\le i

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1987 Paper 2 Q1
D: 1500.0 B: 1500.0
quadratics algebraic identity proof inequality generalisation expanding brackets sum of squares

Prove that:

  1. if \(a+2b+3c=7x\), then \[ a^{2}+b^{2}+c^{2}=\left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2}; \]
  2. if \(2a+3b+3c=11x\), then \[ a^{2}+b^{2}+c^{2}=\left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2}. \]
Give a general result of which \((i) \)and \((ii) \)are special cases.

Solution:

  1. \begin{align*} \left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2} &= x^2-2ax+a^2 + 4x^2 -4bx+b^2 + 9x^2-6cx + c^2 \\ &= (1^2 + 2^2 + 3^2)x^2 - 2x(a+2b+3c) +a^2+b^2 + c^2 \\ &= 14x^2 - 2x(7x) + a^2 + b^2 + c^2 \\ &= a^2 + b^2 + c^2 \end{align*}
  2. \begin{align*} \left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2} &= (2^2+3^2+3^2)x^2 - 2x(2a+3b+3c) + (a^2 + b^2+c^2) \\ &= 22x^2 - 2x(11x) + a^2+b^2+c^2 \\ &= a^2+b^2+c^2 \end{align*}
The general result is: If \(\frac{A^2+B^2+C^2}{2}x =Aa+Bb+Cc\) then \((Ax-a)^2 + (Bx-b)^2 + (Cx-c)^2 = a^2+b^2+c^2\) Alternatively, if \(\lambda = \frac{2\mathbf{x} \cdot \mathbf{y}}{\Vert x \Vert^2}\) then \(\Vert \lambda \mathbf{x} - \mathbf{y}\Vert^2 = \Vert \mathbf{y} \Vert^2\) which is easy to see is true.

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