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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 \(y = 3\).
- 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.
- Use your answer to part (i) to deduce an approximate value for the area of region \(B\).
- 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 \(^ { \text {N } }\)}