Parametric or Inverse Function Area

Find area involving curves defined parametrically, implicitly (e.g., y² = ...), or requiring integration with respect to y by rearranging x = f(y).

3 questions · Standard +0.3

1.08f Area between two curves: using integration
Sort by: Default | Easiest first | Hardest first
CAIE P1 2012 November Q8
8 marks Standard +0.8
8 \includegraphics[max width=\textwidth, alt={}, center]{e69332d0-2e45-4a86-a1f9-5d83bca1ad9b-3_629_853_251_644} The diagram shows the curve \(y ^ { 2 } = 2 x - 1\) and the straight line \(3 y = 2 x - 1\). The curve and straight line intersect at \(x = \frac { 1 } { 2 }\) and \(x = a\), where \(a\) is a constant.
  1. Show that \(a = 5\).
  2. Find, showing all necessary working, the area of the shaded region.
CAIE P1 2016 November Q7
7 marks Standard +0.3
7 \includegraphics[max width=\textwidth, alt={}, center]{9f17f7b8-b54d-467d-be26-21c599ce6ca2-3_704_558_258_790} The diagram shows parts of the curves \(y = ( 2 x - 1 ) ^ { 2 }\) and \(y ^ { 2 } = 1 - 2 x\), intersecting at points \(A\) and \(B\).
  1. State the coordinates of \(A\).
  2. Find, showing all necessary working, the area of the shaded region.
OCR C2 2011 June Q4
7 marks Moderate -0.3
4 \includegraphics[max width=\textwidth, alt={}, center]{4d03f4e3-ae6c-4a0d-ae4d-d89258d2919a-3_588_1136_255_502} The diagram shows the curve \(y = - 1 + \sqrt { x + 4 }\) and the line \(y = 3\).
  1. Show that \(y = - 1 + \sqrt { x + 4 }\) can be rearranged as \(x = y ^ { 2 } + 2 y - 3\).
  2. Hence find by integration the exact area of the shaded region enclosed between the curve, the \(y\)-axis and the line \(y = 3\).