IB Mathematics SL 5.5 integration as anti-differentiation AI HL Paper 2- Exam Style Questions- New Syllabus

Question

Linda owns a field, represented by the shaded region \( R \). The plan view of the field is shown in the following diagram, where both axes represent distance and are measured in metres.

Field Diagram

The segments \([AB]\), \([CD]\) and \([AD]\) respectively represent the western, eastern and southern boundaries of the field. The function, \( f(x) \), models the northern boundary of the field between points \( B \) and \( C \) and is given by
\( f(x) = -\frac{x^2}{50} + 2x + 30 \), for \( 0 \leq x \leq 70 \).

(a) (i) Find \( f'(x) \).
(ii) Hence find the coordinates of the point on the field that is furthest north.

(b) (i) Write down the integral which can be used to find the area of the shaded region \( R \).
(ii) Find the area of Linda’s field.

Linda used the trapezoidal rule with ten intervals to estimate the area. This calculation underestimated the area by \( 11.4 \, \text{m}^2 \).

(c) (i) Calculate the percentage error in Linda’s estimate.
(ii) Suggest how Linda might be able to reduce the error whilst still using the trapezoidal rule.

Linda would like to construct a building on her field. The square foundation of the building, \( EFGH \), will be located such that \([EH]\) is on the southern boundary and point \( F \) is on the northern boundary of the property. A possible location of the foundation of the building is shown in the following diagram.

Additional Diagram

The area of the square foundation will be largest when \([GH]\) lies on \([CD]\).

(d) (i) Find the \( x \)-coordinate of point E for the largest area of the square foundation of building EFGH.
(ii) Find the largest area of the foundation.

▶️ Answer/Explanation

Markscheme

(a)(i)
    Differentiate \( f(x) = -\frac{x^2}{50} + 2x + 30 \)
    \( f'(x) = \frac{d}{dx} \left(-\frac{x^2}{50} + 2x + 30\right) \)
    \( = -\frac{2x}{50} + 2 \)
    Simplify: \( f'(x) = -\frac{x}{25} + 2 \)
Result:
\( f'(x) = \frac{-x}{25} + 2 \)

(a)(ii)
    Set \( f'(x) = 0 \) for maximum: \( -\frac{x}{25} + 2 = 0 \)
    \( \frac{x}{25} = 2 \)
    \( x = 50 \)
    Substitute into \( f(x) \): \( f(50) = -\frac{50^2}{50} + 2 \times 50 + 30 \)
    \( 50^2 = 2500 \), \( \frac{2500}{50} = 50 \)
    \( -50 + 100 + 30 = 80 \)
    So coordinates = \( (50, 80) \)
Result:
\( (50, 80) \)

(b)(i)
Area from \( x = 0 \) to \( x = 70 \) under \( f(x) \)
Integral: \( \int_{0}^{70} \left(-\frac{x^2}{50} + 2x + 30\right) dx \)
Result:
\( \int_{0}^{70} \left(\frac{-x^2}{50} + 2x + 30\right) dx \)

(b)(ii)
    Integrate: \( \int \left(-\frac{x^2}{50} + 2x + 30\right) dx \)
    \( = -\frac{x^3}{150} + x^2 + 30x + C \)
    Evaluate from 0 to 70:
    At \( x = 70 \): \( -\frac{70^3}{150} + 70^2 + 30 \times 70 \)
    \( 70^3 = 343000 \), \( \frac{343000}{150} \approx 2286.6667 \)
    \( 70^2 = 4900 \), \( 30 \times 70 = 2100 \)
    \( -2286.6667 + 4900 + 2100 = 4713.3333 \)
    At \( x = 0 \): 0
    Area = \( 4713.3333 – 0 \approx 4713 \) m²
Result:
\( 4713 \) m²

(c)(i)
    Actual area = 4713 m², underestimate by 11.4 m²
    Estimated area = \( 4713 – 11.4 = 4701.6 \) m²
    Error = \( |4701.6 – 4713| = 11.4 \) m²
    Percentage error = \( \frac{11.4}{4713} \times 100 \approx 0.2419\% \)
Result:
\( 0.242\% \)

(c)(ii)
Reduce the width of the intervals or increase the number of intervals
Result:
Reduce the width of the intervals or increase the number of intervals

(d)(i)
    Square foundation EFGH, \([EH]\) on southern boundary, \( F \) on \( f(x) \), \([GH]\) on \([CD]\)
    Width \( = 70 – x \), length \( = f(x) = -\frac{x^2}{50} + 2x + 30 \)
    Equate sides: \( -\frac{x^2}{50} + 2x + 30 = 70 – x \)
    \( -\frac{x^2}{50} + 3x – 40 = 0 \)
    Multiply by 50: \( -x^2 + 150x – 2000 = 0 \)
    \( x^2 – 150x + 2000 = 0 \)
    Discriminant: \( 150^2 – 4 \times 1 \times 2000 = 22500 – 8000 = 14500 \)
    \( \sqrt{14500} \approx 120.415 \)
    \( x = \frac{150 \pm 120.415}{2} \)
    \( x \approx \frac{150 + 120.415}{2} \approx 135.207 \) or \( \frac{150 – 120.415}{2} \approx 14.792 \)
    Valid in \( 0 \leq x \leq 70 \), so \( x \approx 14.792 \)
Result:
\( 14.8 \) m

(d)(ii)
    Area of square foundation:
    \( (70 – 14.792\ldots)^2 \)
    OR \( (55.2079\ldots)^2 \)
    OR \( \left(-\frac{(14.792\ldots)^2}{50} + 2 \times (14.792\ldots) + 30\right)^2 \)
    Calculate \( 70 – 14.792\ldots = 55.2079\ldots \)
    \( (55.2079\ldots)^2 \approx 3047.92\ldots \) m²
    To 3 s.f.: \( 3050 \) m²
Result:
\( 3050 \) m²

IB Mathematics SL 5.5 integration as anti-differentiation AI HL Paper 2- Exam Style Questions- New Syllabus

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