A school canteen models the cost of ingredients by the curve \(y = 3x^{2} + 2\). The region \(R\) is bounded by the curve, the \(x\)-axis and the lines \(x ...
Question 1 Report
A school canteen models the cost of ingredients by the curve \(y = 3x^{2} + 2\). The region \(R\) is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).
- Write down the value of \(y\) when \(x = 3\). (1)
- Calculate the area of \(R\). (3)
Answer Details
(a) Substituting \(x = 3\) into \(y = 3x^{2} + 2\):
\[y = 3(9) + 2 = 29\][1]
(b) The area between a curve and the \(x\)-axis, from \(x = a\) to \(x = b\), is found by integrating the equation of the curve and evaluating between those limits. The region \(R\) shown runs from \(x = 1\) to \(x = 3\), and the curve lies above the \(x\)-axis throughout, so the integral gives the area directly.
Integrating raises each index by one and divides by the new index:
\[\int (3x^{2} + 2)\,dx = \frac{3x^{3}}{3} + 2x = x^{3} + 2x\][1]
Evaluating at the upper limit and then at the lower limit:
\[\text{at } x = 3: \quad 27 + 6 = 33 \qquad \text{at } x = 1: \quad 1 + 2 = 3\][1]
The area is the upper value minus the lower value:
\[\text{Area} = 33 - 3 = 30\][1]
The constant of integration is not needed for a definite integral, because it appears in both evaluations and cancels in the subtraction.
A rough check supports the answer: the region is \(2\) units wide and the curve rises from \(y = 5\) at \(x = 1\) to \(y = 29\) at \(x = 3\), so an area of \(30\) square units is the right order of size.
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