IBDP Maths SL 5.5 Introduction to integration as anti differentiation of functions AA HL Paper 1- Exam Style Questions- New Syllabus

Question

A rocket moving in a straight line has velocity \(v\) km s⁻¹ and displacement \(s\) km at time \(t\) seconds. The velocity \(v\) is given by \(v(t) = 6e^{2t} + t\). When \(t = 0\), \(s = 10\).

Find an expression for the displacement of the rocket in terms of \(t\).

▶️ Answer/Explanation

Solution

Integrate the velocity function:

\(s(t) = \int (6e^{2t} + t) \, dt = 3e^{2t} + \frac{t^2}{2} + C\) (M1A2A1)

Substitute initial condition \(t=0\), \(s=10\):

\(10 = 3e^{0} + 0 + C \Rightarrow C = 7\) (M1A1)

\(\boxed{s(t) = 3e^{2t} + \frac{t^2}{2} + 7}\) (A1 N6)

Question

The following diagram shows the graph of a function \(f\). There is a local minimum point at \(A\), where \(x > 0\).

Graph of function f with local minimum at A

The derivative of \(f\) is given by \(f'(x) = 3x^2 – 8x – 3\).

a. Find the \(x\)-coordinate of \(A\). [5]

b. The \(y\)-intercept of the graph is at \((0,6)\). Find an expression for \(f(x)\).

The graph of a function \(g\) is obtained by reflecting the graph of \(f\) in the \(y\)-axis, followed by a translation of \(\left(\begin{array}{c}m\\ n\end{array}\right)\). [6]

c. Find the \(x\)-coordinate of the local minimum point on the graph of \(g\). [3]

▶️ Answer/Explanation

Solution a

Set \(f'(x) = 0\): \(3x^2 – 8x – 3 = 0\) (M1)

Solve: \((3x + 1)(x – 3) = 0\) (M1A1)

Solutions: \(x = -\frac{1}{3}\) or \(x = 3\)

Since \(x > 0\): \(\boxed{x = 3}\) (A2 N3)

Solution b

Integrate \(f'(x)\): \(\int (3x^2 – 8x – 3)dx = x^3 – 4x^2 – 3x + C\) (M1A1A1)

Using \((0,6)\): \(6 = 0 – 0 – 0 + C \Rightarrow C = 6\) (A1)

\(\boxed{f(x) = x^3 – 4x^2 – 3x + 6}\) (A1 N6)

Solution c

Reflect \(f\): \(g(x) = f(-x + m) + n\) (A1)

Minimum was at \(x=3\), now at \(x=-3\) after reflection (M1)

After translation: \(\boxed{x = -3 + m}\) (A1 N3)

—Markscheme—

a. [5 marks]

Setting derivative to zero (M1)

Solving quadratic (M1A1)

Correct solution (A2)

b. [6 marks]

Correct integration (3 marks) (M1A1A1)

Finding constant (A1)

Final answer (A1 N6)

c. [3 marks]

Reflection (A1)

Understanding transformation (M1)

Final coordinate (A1 N3)

Total [14 marks]

Question

a. Calculate \(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} \). [6]

b. Find \(\int {{{\tan }^3}x{\text{d}}x} \). [3]

▶️ Answer/Explanation

Solution a

EITHER

Let \(u = \tan x;{\text{ d}}u = {\sec ^2}x{\text{d}}x\) (M1)

Consideration of change of limits (M1)

\(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{1}{{{u^{\frac{1}{3}}}}}{\text{d}}u} \) (A1)

Note: Do not penalize lack of limits.

\( = \left[ {\frac{{3{u^{\frac{2}{3}}}}}{2}} \right]_1^{\sqrt 3 }\) A1

\( = \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} – \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} – 3}}{2}} \right)\) A1A1 N0

\(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \left[ {\frac{{3{{(\tan x)}^{\frac{2}{3}}}}}{2}} \right]_{\frac{\pi }{4}}^{\frac{\pi }{3}}\) M2A2

\( = \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} – \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} – 3}}{2}} \right)\) A1A1 N0

\(\boxed{\frac{3\sqrt[3]{3} – 3}{2}}\)

Solution b

\(\int {{{\tan }^3}x{\text{d}}x} = \int {\tan x({{\sec }^2}x – 1){\text{d}}x} \) M1

\( = \int {(\tan x \times {{\sec }^2}x – \tan x){\text{d}}x} \)

\( = \frac{1}{2}{\tan ^2}x – \ln \left| {\sec x} \right| + C\) A1A1

Note: Do not penalize the absence of absolute value or C.

\(\boxed{\frac{1}{2}\tan^2x – \ln|\sec x| + C}\)

IBDP Maths SL 5.5 Introduction to integration as anti differentiation of functions AA HL Paper 1- Exam Style Questions

IBDP Maths SL 5.5 Introduction to integration as anti differentiation of functions AA HL Paper 1- Exam Style Questions- New Syllabus

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