IBDP Maths SL 5.11 Definite integrals AA HL Paper 2- Exam Style Questions- New Syllabus
Let \( f(x) = \frac{1}{4}x^2 + 2 \). The line \( L \) is the tangent to the curve of \( f \) at (4, 6).
Let \( g(x) = \frac{90}{3x + 4} \), for \( 2 \le x \le 12 \). The following diagram shows the graph of \( g \).
(a) Find the values of the constants for the equation of \( L \) [4]
(b) Find the values of the constants for the area of the region enclosed by the curve of \( g \), the \( x \)-axis, and the lines \( x = 2 \) and \( x = 12 \). Give your answer in the form \( a \ln b \), where \( a, b \in \mathbb{Z} \) [6]
(c) Find the values of the constants for the area enclosed by the lines \( L \), \( x = 2 \), \( x = 12 \), and the graph of \( h \), where the graph of \( g \) is reflected in the \( x \)-axis to give the graph of \( h \), and the area enclosed by \( L \), \( x = 2 \), \( x = 12 \), and the \( x \)-axis is 120 \( \text{cm}^2 \) [3]
▶️ Answer/Explanation
[13 marks]
(a) \( f'(x) = \frac{1}{2}x \) (A1).
\( f'(4) = 2 \) (M1, A1).
Equation \( y – 6 = 2(x – 4) \) or \( y = 2x – 2 \) (A1).
(b) \( \text{area} = \int_2^{12} \frac{90}{3x + 4} \, dx \) (A1, A1).
\( = 30 \ln (3x + 4) \big|_2^{12} \) (M1).
\( = 30 \ln (3 \cdot 12 + 4) – 30 \ln (3 \cdot 2 + 4) = 30 \ln 40 – 30 \ln 10 \) (A1).
\( = 30 \ln \frac{40}{10} = 30 \ln 4 \) (A1, A1).
(c) Area approach: \( \text{area} = 120 + 30 \ln 4 \) (M1, A2).
Markscheme Answers:
(a) \( f'(x) = \frac{1}{2}x \) (A1), \( f'(4) = 2 \) (M1, A1), \( y – 6 = 2(x – 4) \) or \( y = 2x – 2 \) (A1)
(b) \( \int_2^{12} \frac{90}{3x + 4} \, dx \) (A1, A1), substituting limits (M1), correct working (A1), \( 30 \ln \frac{40}{10} \) (A1), \( 30 \ln 4 \) (A1)
(c) Valid approach (M1), \( 120 + 30 \ln 4 \) (A2)
[Total 13 marks]
(a) Find the values of the constants for \( \int_4^{10} (x – 4) \, dx \) [4]
(b) Find the values of the constants for the volume of the solid formed when the shaded region \( R \), enclosed by the graph of \( f(x) = \sqrt{x – 4} \) for \( x \ge 4 \), the line \( x = 10 \), and the \( x \)-axis, is rotated 360° about the \( x \)-axis [3]
▶️ Answer/Explanation
[7 marks]
(a) \( \int (x – 4) \, dx = \frac{x^2}{2} – 4x \) (A1, A1).
\( \left[ \frac{x^2}{2} – 4x \right]_4^{10} = (\frac{10^2}{2} – 4 \cdot 10) – (\frac{4^2}{2} – 4 \cdot 4) \) (M1).
\( = (50 – 40) – (8 – 16) = 10 – (-8) = 18 \) (A1).
(b) \( \text{volume} = \pi \int_4^{10} (\sqrt{x – 4})^2 \, dx \) (M1).
\( = \pi \int_4^{10} (x – 4) \, dx = \pi \cdot 18 = 18\pi \) (A1, A1).
Markscheme Answers:
(a) \( \frac{x^2}{2} – 4x \) or \( \frac{(x – 4)^2}{2} \) (A1, A1), substituting limits (M1), \( 18 \) (A1)
(b) \( \pi \int_4^{10} f(x)^2 \, dx \) (M1), \( \pi \int_4^{10} (x – 4) \, dx \) (A1), \( 18\pi \) (A1)
[Total 7 marks]