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\includegraphics[max width=\textwidth, alt={}, center]{9e95415c-00f5-4b52-a443-0b946602b3b4-4_387_624_287_717} The diagram shows part of the curve \(y = - 3 + 2 \sqrt { x + 4 }\). The point \(P ( 5,3 )\) lies on the curve. Region \(A\) is bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 5\). Region \(B\) is bounded by the curve, the \(y\)-axis and the line \(y = 3\).

1. Use the trapezium rule, with 2 strips each of width 2.5 , to find an approximate value for the area of region \(A\), giving your answer correct to 3 significant figures.
2. Use your answer to part (i) to deduce an approximate value for the area of region \(B\).
3. By first writing the equation of the curve in the form \(x = \mathrm { f } ( y )\), use integration to show that the exact area of region \(B\) is \(\frac { 14 } { 3 }\). \section*{END OF QUESTION PAPER} \section*{OCR \(^ { \text {N } }\)}