IB Mathematics SL 5.5 integration as anti-differentiation AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Consider the cubic function \( f(x) = x^3 – 9x^2 + 23x – 15 \), defined over the domain \( 0 \leq x \leq 5 \).
The graph of \( y = f(x) \) intersects the \( x \)-axis at three distinct points: \( x = 1 \), \( x = 5 \), and \( x = a \).
(a) Determine the value of \( a \).
(b) Using your graphic display calculator (GDC), calculate the area of the region completely enclosed by the curve and the \( x \)-axis between the boundaries \( x = 1 \) and \( x = a \).
Most appropriate topic codes (IB Mathematics: applications and interpretation):
• SL 2.5: Modelling with cubic functions of the form \( f(x) = ax^3 + bx^2 + cx + d \) — part (a)
• SL 5.5: Area of a region enclosed by a curve \( y = f(x) \) and the \( x \)-axis — part (b)
• SL 5.5: Values of definite integrals using technology — part (b)
• SL 5.5: Area of a region enclosed by a curve \( y = f(x) \) and the \( x \)-axis — part (b)
• SL 5.5: Values of definite integrals using technology — part (b)
▶️ Answer/Explanation
(a)
Since \( x = 1 \) and \( x = 5 \) are roots, \( (x-1)(x-5) = x^2 – 6x + 5 \) is a factor.
Divide \( f(x) \) by \( x^2 – 6x + 5 \):
\( \frac{x^3 – 9x^2 + 23x – 15}{x^2 – 6x + 5} = x – 3 \)
So \( f(x) = (x-1)(x-5)(x-3) \).
The third root is \( a = 3 \).
\(\boxed{3}\)
(b)
Between \( x = 1 \) and \( x = 3 \), \( f(x) \geq 0 \), so the area is:
\( \text{Area} = \int_{1}^{3} \bigl( x^3 – 9x^2 + 23x – 15 \bigr) \, dx \)
Using a GDC (or by integration):
\( \int_{1}^{3} (x^3 – 9x^2 + 23x – 15) \, dx = 4 \)
\(\boxed{4}\)