17M.2.AHL.TZ1.H_4

pestleMathematicsAIHLPaper 217M· ahl-5-12-areas-under-a-curve-onto-x-or-y-axis-volumes-of-revolution-about-x-and-ysource ↗

The region $A$ is enclosed by the graph of $y = 2\arcsin (x - 1) - \frac{\pi }{4}$ , the $y$ -axis and the line $y = \frac{\pi }{4}$ .

Write down a definite integral to represent the area of $A$ .

[4]

Calculate the area of $A$ .

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$2\arcsin (x - 1) - \frac{\pi }{4} = \frac{\pi }{4}$ (M1)

$x = 1 + \frac{1}{{\sqrt 2 }}\,\,\,( = 1.707 \ldots )$ (A1)

$\int\limits_0^{1 + \frac{1}{{\sqrt 2 }}} {\frac{\pi }{4} - \left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right)dx}$ M1A1

Note: Award M1 for an attempt to find the difference between two functions, A1 for all correct.

METHOD 2

when $x = 0,{\text{ }}y = \frac{{ - 5\pi }}{4}\,\,\,( = - 3.93)$ A1

$x = 1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)$ M1A1

Note: Award M1 for an attempt to find the inverse function.

$\int_{\frac{{ - 5\pi }}{4}}^{\frac{\pi }{4}} {\left( {1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)} \right){\text{d}}y}$ A1

METHOD 3

$\int_0^{1.38...} {\left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right){\text{d}}x} \left| + \right.\int\limits_0^{1.71...} {\frac{\pi }{4}{\text{d}}x - \int\limits_{1.38...}^{1.71...} {\left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right)dx} }$ M1A1A1A1

Note: Award M1 for considering the area below the $x$ -axis and above the $x$ -axis and A1 for each correct integral.

[4 marks]

${\text{area}} = 3.30{\text{ (square units)}}$ A2

[2 marks]

Examiners’ report

[N/A]