IBDP Maths SL 5.11 Definite integrals AA HL Paper 1- Exam Style Questions- New Syllabus

A line with equation \(y=-3x+9\) intersects the axes at the points \(P(0,9)\) and \(Q(3,0)\).

A parabola of the form \(y=ax^2+c\), where \(a,c\in\mathbb{Z}\), also passes through the points \(P\) and \(Q\).

This is shown in the following diagram.

(a)

(i) Write down the value of \(c\).

(ii) Find the value of \(a\). [3]

The region enclosed by the line and the parabola is shaded in the following diagram.

(b) Find the area of the shaded region. [4]

Most-appropriate topic codes (IB DP Mathematics: Analysis and Approaches):

• TOPIC SL 2.6 Quadratic functions, their graphs and determining parameters from given points. (Part a)

• TOPIC SL 5.11 Definite integrals and areas between curves. (Part b)

▶️ Answer/Explanation

(a)(i)
The parabola \(y=ax^2+c\) passes through \(P(0,9)\).

Substitute \(x=0\) and \(y=9\).

\(9=a(0)^2+c\)

\(c=9\)

✅ Answer: \(c=9\)

(a)(ii)
The parabola also passes through \(Q(3,0)\).

Substitute \(x=3\), \(y=0\) and \(c=9\) into \(y=ax^2+c\).

\(0=a(3)^2+9\)

\(0=9a+9\)

\(9a=-9\)

\(a=-1\)

Therefore, the equation of the parabola is \(y=9-x^2\).

✅ Answer: \(a=-1\)

(b)
Between \(x=0\) and \(x=3\), the parabola \(y=9-x^2\) lies above the line \(y=9-3x\).

The enclosed area is therefore found by subtracting the equation of the line from the equation of the parabola.

\(\text{Area}=\displaystyle\int_0^3\left[(9-x^2)-(9-3x)\right]\,dx\)

\(\text{Area}=\displaystyle\int_0^3(3x-x^2)\,dx\)

\(\text{Area}=\left[\dfrac{3}{2}x^2-\dfrac{1}{3}x^3\right]_0^3\)

\(\text{Area}=\left[\dfrac{3}{2}(3)^2-\dfrac{1}{3}(3)^3\right]-0\)

\(\text{Area}=\dfrac{27}{2}-9\)

\(\text{Area}=\dfrac{27}{2}-\dfrac{18}{2}=\dfrac{9}{2}\)

✅ Answer: \(\dfrac{9}{2}\) square units

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